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Okay, so the idea with this is to compare an ideal binary mixture with a nonideal binary mixture wherein negative deviation occurs upon creation of the mixture.


"SATURATED" VAPOR PRESSURE

If you remember Raoult's law, it dealt with the following relationship:

#\mathbf(P_j = chi_jP_j^@)#

where:

  • #P_j# is the vapor pressure of the solution containing solute (i.e. non-pure).
  • #chi_j = (n_j)/(n_i + n_j)# is the mol fraction of solvent #j# in solution. When #chi_j darr#, #chi_j < 1#, and #chi_i#, the mol fraction of solute #i#, is increasing.
  • #P_j^@# is the vapor pressure of the solution containing NO solute at all (i.e. purely solvent). This is known as the vapor pressure of the pure solvent, or the "saturated" vapor pressure.

Let's just do the derivation part of the problem right here. Starting from Dalton's law of partial pressures, we just have, for the ideal binary mixture #AB#:

#P_"tot" = color(blue)(P_(AB)^@) = P_A + P_B#

#= chi_AP_A^@ + chi_BP_B^@#

#= chi_AP_A^@ + (1-chi_A)P_B^@#

#= chi_AP_A^@ + P_B^@ - chi_AP_B^@#

#= color(blue)(P_B^@ + chi_A(P_A^@ - P_B^@))#

The physical interpretation of this result is that the total pressure can be computed for the mixture of any two substances, ideal or not, by using their pure vapor pressures and by knowing how much #A# you add to #B#.

If you look back at how I did this derivation, you should notice that #chi_B + chi_A = 1#. That makes sense though, since the mol fractions of the only two components of the ideal binary mixture should add up to #100%# of the components in the mixture.

Thus, you should remember that #chi_A = 1-chi_B#.

ENERGIES OF INTERACTION IN AN IDEAL BINARY MIXTURE

For the ideal binary mixture #AB#, we say that the intermolecular force of attraction between liquid #A# and liquid #B# are equal to those between #A# with itself and #B# with itself. This can be expressed as:

#2epsilon_(AB) = epsilon_(A A) + epsilon_(BB)#

This says that the energy of the system after mixing #A# and #B# with each other is equal to the energy of the system before mixing #A# and #B# with each other.

POSITIVE AND NEGATIVE DEVIATIONS FROM IDEAL MOLAR VOLUMES

Now, we can have two variations on this.


#\mathbf(2epsilon_(AB) < epsilon_(A A) + epsilon_(BB))# (1)

Physical Chemistry: A Molecular Approach, McQuarrie

In (1), the energy #epsilon# for the interaction between #A# and #B# with themselves is greater than that for the interaction between #A# and #B# with each other.

That means the particles, after mixing, have a lower overall energy, and thus, mixing is favorable. Thus, the average distance between #A# and #B# is smaller, and we have what's called negative deviation---the volume of the solution is smaller than for the ideal solution.

That leads to a dip relative to Raoult's law behavior when the mol fraction of the solvent is less than #1# but that of the solute is not yet #1#.


#\mathbf(2epsilon_(AB) > epsilon_(A A) + epsilon_(BB))# (2)

Physical Chemistry: A Molecular Approach, McQuarrie

Similarly, in (2), since the energy for the #AB# interaction is greater than for the #A A# and #BB# interactions combined, mixing is unfavorable, and the average distance between particles is larger; thus, the volume increases and we have positive deviation.

That leads to a "hump" relative to Raoult's law behavior when the mol fraction of the solvent is less than #1# but that of the solute is not yet #1#.

THE AB VS THE RS SOLUTION

Based on your problem setup, what you're dealing with is ideality for the AB solution and negative deviation for the RS solution.


For the ideal mixture, the total vapor pressure, #P_(AB)^@#, of the solution should increase linearly with the addition of solute #A# (and thus increase #P_A^@#) into a solution that contained only #\mathbf(B)# at first.

This is because there is no favorability in either the positive or negative direction when it comes to mixing #A# and #B#; the resultant energy of their combination, in any combination (#A A#, #BB#, or #AB#), doesn't make a difference.

Note that since we defined #\mathbf(P_A^@ > P_B^@)#, it makes sense that the total vapor pressure increases as more #A# is added to the solution.


For the #RS# mixture, we noted that it had negative deviation. This manifests itself as a dip downwards in the graph of total vapor pressure, #P_(RS)#, as you add #R# (and thus an increase in #P_R^@#), and an increase again to #P_(RS)# as usual as you add more #R#.

When there is #50%# #R# and #50%# #S#, or whatever particular quantity of #R#, the solution is most favorably mixed and the volume contracts relative to ideal (Raoult's law) behavior.

When there is #100%# #R#, note that we defined #P_A^@ > P_B^@#, #P_R^@ = P_A^@#, and #P_S^@ = P_B^@#.

Thus, #\mathbf(P_R^@ > P_S^@)#, and it makes sense that ultimately, with #100%# #R#, #P_R^@ > P_S^@#.

Answer:

Here's what I got.

Explanation:

Based on the answer options given to you, it appears as though the question is incomplete.

No mention of lead's possible reaction was made in the question, yet lead is mentioned in the answers, so I can only assume that you're missing some information.

Now, you can answer the question intuitively. You know that when a zinc rod is placed in a tin(II) chloride, #"SnCl"_2#, aqueous solution, tin will precipitate on the zinc rod and zinc will go into solution as zinc cations, #"Zn"^(2+)#.

Even without writing a balanced chemical equation, you can look at the info given and say that because zinc is losing electrons to form zinc cations, that must mean that it's being oxidized.

Likewise, tin is going from a #2+# oxidation state in tin(II) chloride to being precipitated as tin metal. This means that it's gaining electrons, which must mean that it's being reduced.

So zinc acts as a reducing agent because it reduces tin(II) cations to tin metal. In other words, the tin(II) cations act as an oxidizing agent because they oxidize zinc metal to zinc cations.

Now, I think that the second part of the question features a lead rod being placed in a tin(II) chloride aqueous solution. Moreover, judging from the options given to you, I'd say that when this happens, no reaction takes place.

In other words, lead metal is unable to reduce tin(II) cations to tin metal, or tin(II) cations are unable to oxidize lead metal to lead(II) cations, #"Pb"^(2+)#.

If this is the case, you can say that

  • Option (A) is #color(red)("incorrect")# because lead metal is actually a weaker reducing agent than zinc metal.

  • Option (B) is also #color(red)("incorrect")# because tin does not act as a reducing agent when paired with zinc, it acts as an oxidizing agent.

  • Option (C) is #color(green)("correct")# because zinc manages to reduce the tin(II) cations to tin metal but lead does not, and so zinc is indeed a stronger reducing agent than lead.

  • Option (D) is #color(red)("incorrect")# because zinc does not act as an oxidizing agent in the reaction, so saying that tin is a stronger oxidizing agent than a reducing agent doesn't really make sense to me.

  • Option (E) is also #color(red)("incorrect")# because zinc is oxidized to zinc cations in solution, whereas lead is not. This implies that the lead(II) cations are actually stronger oxidizing agents than the zinc cations.

You can write out the net ionic equation for the reaction between zinc and aqueous tin(II) chloride

#"Zn"_ ((s)) + "Sn"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Sn"_ ((s))#

Looking at this reaction, you can say that you have

#"Zn"_ ((s)) -># a stronger reducing agent is being converted to #"Zn"_ ((aq))^(2+)#, a weaker oxidizing agent

#"Sn"_ ((aq))^(2+)-># a stronger oxidizing agent is being converted to #"Sn" _((s))#, a weaker reducing agent

For lead you would have

#"Pb"_ ((s)) + "Sn"_ ((aq))^(2+) -> "N.R."#

This means that you have

#"Pb"_ ((s)) -># a weaker reducing agent is not being converted to a stronger oxidizing agent

#"Sn"_ ((aq))^(2+) -># a weaker oxidizing agent is not being converted to a stronger reducing agent

Answer:

#T_"f sol" = -1.16^@"C"#

Explanation:

When dealing with the freezing point of a solution, your first goal is to determine its molality by

  • determining the number of moles of solute present in the solution
  • determining the mass of the solvent in kilograms

Once you know the molality of the solution, you can find the freezing-point depression by using the equation

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#, where

#DeltaT_f# - the freezing-point depression;
#i# - the van't Hoff factor
#K_f# - the cryoscopic constant of the solvent;
#b# - the molality of the solution.

In your case, the cryoscopic constant of water is said to be

#K_f = 1.86^@"C kg mol"^(-1)#

Glucose is a non-electrolyte, i.e. it remains undissociated in aqueous solution, which means that its van't Hoff factor will be #i=1#.

So, you know that your solution contains #"53.4 g"# of glucose. To determine how many moles of glucose it contains, use the compound's molar mass

#53.4 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))) = "0.2964 moles glucose"#

The solution contains #"475 g"# of water, which is the solvent. Convert this to kilograms

#475color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.475 kg"#

The molaity of the solution, which is defined as

#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#

will thus be

#b = "0.2964 moles"/"0.475 kg" = "0.624 mol kg"^(-1)#

Now plug in your values and solve for the freezing-point depression

#DeltaT_f = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.624 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 1.161^@"C"#

The freezing-point depression is defined as

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#, where

#T_f^@# - the freezing point of the pure solvent
#T_"f sol"# - the freezing point of the solution

This means that the freezing point of the solution will be

#T_"f sol" = T_f^@ - DeltaT_f#

#T_"f sol" = 0^@"C" - 1.161^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.16^@"C")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Answer:

#50%#

Explanation:

The idea here is that you need to use the fact that deuterium, #"D"#, which is one of the three naturally occurring isotopes of hydrogen, #"H"#, contains two nucleons in its nucleus.

#color(white)()#

http://www.qlarivia.com/what-is-deuterium/

As you know, the relative atomic mass of an atom, #A_r#, is calculated by dividing the atomic mass of the atom, #m_a#, by the mass of #1/12"th"# of an atom of #""^12"C"#, which is equivalent to the mass of one nucleon, i.e. one proton or one neutron.

The relative atomic mass of hydrogen is approximately equal to #1# because its nucleus contains one proton. Since the nucleus of a deuterium atom contains a proton and a neutron, its relative atomic mass will be equal to #2#.

Now, water, #"H"_2"O"#, contains two hydrogen atoms and one oxygen atom. Heavy water, #"D"_2"O"#, contains two deuterium atoms and one oxygen atom.

This means that the difference between the relative molecular mass of water and the relative molecular mass of heavy water will be equal to twice the difference between #A_r# of hydrogen and #A_r# of deuterium.

The relative atomic mass of oxygen is #16#, which means that the relative molecular masses of water and of heavy water are

#A_("r H"_2"O") = 2 xx 1 + 16 = 18#

#A_("r D"_2"O") = 2 xx 2 + 16 = 20#

Your sample has an average relative molecular mass of #19#, which can only mean that it contains equal amounts of water and of heavy water.

Mathematically, you can show this by using a system of two equations that have #x# as the decimal abundance of #"H"_2"O"# and #y# as the decimal abundance of #"D"_2"O"#

#{(19 = x * 18 + y * 20), (color(white)(9)1 = x + y ) :}#

This will get you

#19 =18x + 20 - 20x#

#2x = 1 implies x = 1/2#

This means that

#y = 1 - 1/2 = 1/2#

and thus your sample contains #50% "H"_2"O"# and #50% "D"_2"O"#.

If you already found the oxidation half-reaction, then you should have already found the reduction half-reaction. That's why they're called half-reactions. Once you have one of them, you know the other is the other kind.

So, I assume you haven't found the oxidation half-reaction either.

Reduction

#"Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) + 6e^(-) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)#

Oxidation

#2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq) + 6e^(-)#

Ultimately I got the full reaction as:

#"Cr"_2"O"_7^(2-)(aq) + 2"NH"_4^(+)(aq) -> "Cr"_2"O"_3(s) + "N"_2(g) + 4"H"_2"O"(l)#


We have two half-reactions corresponding to each species (#"Cr"#, #"N"# species):

#"Cr"_2"O"_7^(2-)(aq) -> "Cr"_2"O"_3(s)#

#"NH"_4^(+)(aq) -> "N"_2(g)#

Note that in order for #"NH"_4^(+)# to exist in solution (#"pKa" ~~ 9.4#), the solution should be acidic (otherwise, it would exist as #"NH"_3#, above approximately pH #9.4#), so we are balancing in acidic conditions.

REDUCTION HALF-REACTION

The natural oxidation state of #"O"# ion not in a peroxide is #-2#, so by finding the charge balance, you can find the oxidation state of each #"Cr"# ion.

Work this out and convince yourself it works:

  • #"Cr"_2"O"_7^(2-)#: #2x + 7xx(-2) = -2 => color(red)(x = +6)#.
  • #"Cr"_2"O"_3#: #2x + 3xx(-2) = 0 => color(red)(x = +3)#.

#stackrel(color(red)(+6))("Cr"_2)stackrel(color(red)(-2))("O"_7^(2-))(aq) -> stackrel(color(red)(+3))("Cr"_2)stackrel(color(red)(-2))("O"_3)(s)#

Two ways you could understand that this is a reduction half-reaction:

  1. When the number of oxygens increases, the pertinent species is oxidized. So, when the opposite happens, the pertinent species is reduced. Specifically, #+6 -> +3#.
  2. When the oxidation state gets less positive (or more negative), the element was reduced. Specifically, #+6 -> +3#.

Now, let's balance this in acid. Here's how I would do it:

  1. Add whole-number coefficients to balance main species (i.e. #"Cr"#, #"N"#, etc), if needed.
  2. Add water to balance oxygens, if needed.
  3. Add #\mathbf("H"^(+))# to balance hydrogens, if needed.
  4. Add electrons to balance the charges, if needed.

#=> "Cr"_2"O"_7^(2-)(aq) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)#

#=> "Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)#

#=> color(blue)("Cr"_2"O"_7^(2-)(aq) + 8"H"^(+)(aq) + 6e^(-) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l))#

Charge Balance check:

#color(green)((-2) + 8xx(+1) + (-6) = 0)#

OXIDATION HALF-REACTION

Let's identify the oxidation states. Here's this one, unbalanced.

#stackrel(color(red)(-3))("N")stackrel(color(red)(+1))("H"_4^(+)(aq)) -> stackrel(color(red)(0))("N"_2(g))#

  • A compound in its elemental state, like #"N"_2(g)# or #"O"_2(g)#, has an oxidation state of #color(red)(0)#.
  • The more electronegative element in #"NH"_4^(+)# has the negative oxidation state.

So, #"H"# doesn't have an oxidation state of #-1#, but #color(red)(+1)#. Therefore, #+1xx3 - 3 = 0#, and #"N"# has an oxidation state of #color(red)(-3)#.

Since we had #stackrel(color(red)(-3))("N") -> stackrel(color(red)(0))("N")#, nitrogen was oxidized, i.e. its oxidation state became more positive (or less negative).

Next, let's balance this in acid. Same strategy as before, but the nitrogens are unbalanced and we don't need to add water since there is no oxygen on either side.

#=> 2"NH"_4^(+)(aq) -> "N"_2(g)#

#=> 2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq)#

#=> color(blue)(2"NH"_4^(+)(aq) -> "N"_2(g) + 8"H"^(+)(aq) + 6e^(-))#

Charge Balance check:

#color(green)(2xx(+1) = 8xx(+1) + (-6))#

OVERALL BALANCED REACTION

Thus, the overall reaction is...

#"Cr"_2"O"_7^(2-)(aq) + cancel(8"H"^(+)(aq)) + cancel(6e^(-)) -> "Cr"_2"O"_3(s) + 4"H"_2"O"(l)#
#2"NH"_4^(+)(aq) -> "N"_2(g) + cancel(8"H"^(+)(aq)) + cancel(6e^(-))#
#"------------------------------------------------------------------"#
#\mathbf(color(blue)("Cr"_2"O"_7^(2-)(aq) + 2"NH"_4^(+)(aq) -> "Cr"_2"O"_3(s) + "N"_2(g) + 4"H"_2"O"(l)))#

As an overall observation, there is no data missing from the table---it's on purpose. All the information you need is in the question information.

And the best way to learn kinetics is to have an answer key and to try practice problems. It's not easy, so you'll need to put in the practice.

Some general tips:

  • Make sure you know the difference between the rate law for the reaction, the rate of reaction #r(t)#, the rate constant #k#, and the rate of disappearance of reactant #-(d[R])/(dt)# or rate of appearance of product #(d[P])/(dt)#. These can often feel similar but they're not the same.
  • When determining reaction order, don't be afraid to make up numbers and seeing what happens to #r(t)# when you keep one initial concentration constant and change the other one.

The point is to see relationships between changes in rate and changes in reactant concentration.

  • Make sure you know how to write a rate law inside and out; it's often a useful equation to know how to work with, and is central to essentially any kinetics problem.

DISCLAIMER: LONG ANSWER!

(1) WRITING A RATE LAW

The rate law for the reaction is in general written as:

#\mathbf(r(t) = k[X]^m[Y]^n)#

#= \mathbf(-1/(nu_X)(d[X])/(dt) = -1/(nu_Y)(d[Y])/(dt) = 1/(nu_Z)(d[Z])/(dt))#

for the reaction

#nu_X X + nu_Y Y -> nu_Z Z#

where:

  • #r(t)# is the rate as a function of time.
  • #nu# is the stoichiometric coefficient.
  • #k# is the rate constant, in units of #1/("M"^(-1 + m + n)cdot"s")# and #"1 M" = "1 mol/dm"^3#.
  • #pm(d["compound"])/(dt)# is the rate of disappearance of reactant (negative) or appearance of product (positive).
  • #m# and #n# are the orders of each reactant---having no relevance to the stoichiometry of the reaction at all.

For this, #nu_X = nu_Y = nu_Z = 1#, and we need to find #m# and #n#. We don't really need #k#, apparently.

(2) FINDING THE ORDERS OF EACH REACTANT

For the reaction

#"X"(aq) + "Y"(aq) -> "Z"(aq),#

we are basically looking at what happens when we change the initial concentrations of each reactant while keeping the other constant.

We're "rigging" the reaction to see what the orders (contributions) of each reactant are with respect to the rate.

The order for each reactant is found like so:

  1. Look at two trials, where one reactant's initial concentration is kept the same for both trials.
  2. Now look at the other reactant. What happened to its initial concentration?
  3. Now inspect the rate in #"mol/dm"^3cdot"s"#. What happened to the rate?
  4. Based on that, you should be able to find the order.

For trials 1 and 2, halving #[X]# halved the rate. Looking back at the rate law, basically, we are saying:

#[X]^m -> (([X])/2)^m# with #[Y]^n -> [Y]^n# causes #r(t) -> (r(t))/2#.

Therefore #m = 1#; that is, the order of #X# is #color(blue)(1)#.

For trials 2 and 3, doubling #[Y]# quadrupled #r(t)#. Basically, we are saying:

#[Y]^n -> (2[Y])^n# with #[X]^m -> [X]^m# causes #r(t) -> 4r(t)#.

So, #n = 2#, because #2^2 = 4#; that is, the order of #Y# is #color(blue)(2)#.

(3) FINDING A NEW RATE BASED ON NEW CONCENTRATIONS

We already got the information we needed for this in part 2.

Since we know the order of #Y# is #2#, compare to trial 3; we doubled #[Y]#, so the rate quadruples, just like going from trial 2 to 3.

Therefore, without doing much work, #color(blue)(r(t) = "0.016 mol/dm"^3cdot"s")#.

(4) ROLE OF D

Considering trials 4 and 5, the only trials to use #D#, the rate multiplied by #5#, for both of those two concentrations of #D#, relative to trial 3.

That tells me it's a catalyst, because a catalyst changes the mechanism of the reaction in some way, making it faster.

A common way it does so is by lowering the activation energy.

But it's surprising that the rate was the same for trials 4 and 5. I guess it means...

(5) REACTION COORDINATE DIAGRAMS

For this, that is asking you to use your answer in part 4 to draw a reaction coordinate diagram in the presence and absence of #D#. Recall that it is basically a plot of energy vs. the progress of the reaction.

As I said earlier, the activation energy, #E_a# is lower, and additionally, the mechanism is probably different.

For this diagram, #E_a# is the activation energy without #D#, and #E_a^"*"# is with #D#. The dip in the curve is an intermediate formed as a result of using #D#, and we don't need to know what it actually is.

I made up the curve, but it illustrates the idea: #\mathbf(E_a)# is lowered by #D#, the catalyst, speeding up the rate of reaction.

The products are lower in energy, indicating an energetically favorable reaction. That's an assumption, which is just saying that the reaction should happen.

(6) COMPARING TRIALS 3 AND 6

As for trial 6, we know that in comparison to trial 3, the reaction conditions are pretty much the same, except no #D# was added for trial 6. My concern is that it's not completely true.

Everything looks similar, but remember, trials 4 and 5 had to have happened right before trial 6.

So, I would expect that some catalyst #D# was not fully removed from solution.

That should make sense because #D# is aqueous, so it must have been soluble, and soluble catalysts are harder to remove than insoluble catalysts. You can just scoop out a solid catalyst. You can't just do that for a dissolved catalyst!

So, probably, some residual #D# is still there, and the reaction is still faster than it would be without #D#, but not as fast as it would be with more #D#.

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