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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:

The mass defect is #9.141xx10^{-25}" g/atom"#.

Explanation:

Mass defect is the amount of mass that is lost when protons and neutrons combine to form a nucleus. The protons and neutrons become bound to each other, and the binding energy that's released shows up as mass being lost because of the relation #E=mc^2#. So when we talk about mass defect we really mean binding energy.

Nickel-60 has a mass number of 60 and an atomic number of 28, thus 28 protons and 32 neutrons are bound together. The mass of the geee particles is given by:

#(28xx1.00728)+(32xx1.00867)=60.48128" g/mol"#

Compare that with the given atomic mass of nickel-60 #=59.9308" g/mol"#. Take the didference and round to a multiple of #0.0001# matching the given accuracy of the nuckel atomic mass:

Mass defect = #60.48128-59.9308=0.5505" g/mol"#

Note the units. To get grams per atom divide by Avogadro's Number:

#{0.5505" g/mol"}/{6.022xx10^{23}" atoms/mol"}=9.141xx10^{-25}" g/atom"#.

Go back to the molar basis and see how much energy this is in #"J/mol"#. One joule is #1000" g"# mass times #1"m/s"^2# acceleration times #1" m"# distance:

#E=mc^2={0.5505" g/mol"xx(299792458" m/s")^2xx1" J"}/{1000" g m"^2/"s"^2}=4.948xx10^{13}" J/mol"#

This is tremendously larger than the energy changes associated in chemical reactions. This shows the potential power of the nuclear force and processes based on it.

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)))#

The given reversible gaseous reaction :

#2X_2Y(g)" "rightleftharpoons" "2X_2(g)" "+" "Y_2(g)#

#color(red)(I)" "1" "mol" " 0" "mol" " 0" " "mol#

#color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha/2" " "mol#

#color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha/2" " "mol#

Where #alpha# is the degree of dissociation of #X_2Y(g)#

Volume of the vessel being 1 L
the concentrations of the reactants and products at equilibrium will be as follows

#[X_2Y(g)]=(1-alpha)M#

#[X_2(g)]=alphaM#

#[Y_2(g)]=alpha/2M#

Now concentration equilibrium constant

#K_c=([X_2(g)]^2[Y_2(g)])/[X_2Y(g)]^2#

#=>8*10^-6=((alpha)^2*alpha/2)/(1-alpha)^2#

Neglecting #color(red)(alpha)# comparing with 1

#=>alpha^3/2=8*10^-6#

#=>alpha=16^(1/3)*10^-2~~2.5*10^-2-># No. of moles dissociated per mole of reactant

So the percent dissociation #=2.5*10^-2*100=2.5%#

Answer:

#Delta_"pH" = -0.026#

Explanation:

You're adding hydrochloric acid, #"HCl"#, a strong acid, to your buffer, so right from the start you should expect it pH to decrease.

This implies that the change in pH will be negative

#Delta_ "pH" = "pH"_ "final" - "pH"_ "initial" <0#

However, the fact that you're dealing with a buffer solution lets you know that this change will not be significant since the role of a buffer is to resist significant changes in pH that result from the addition of strong acid or strong bases.

So, hydrochloric acid will react with ammonia, #"NH"_3#, a weak base, to produce the ammonium cation, #"NH"_4^(+)#, the ammonia's conjugate acid, and water.

#"HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#

The reaction consumes hydrochloric acid and ammonia in a #1:1# mole ratio. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, one mole of ammonium cations is produced. Keep this in mind.

Use the molarity of the ammonia solution and the volume of the buffer to calculate how many moles it contains

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

#= "0.0100 moles NH"_3#

Do the same for the ammonium cations, #"NH"_4^(+)#

#n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#

#= "0.0100 moles NH"_4^(+)#

Calculate how many moles of hydrochloric acid are being added to the buffer

#n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))#

#= "0.000300 moles HCl"#

You know that hydrochloric acid and ammonia react in a #1:1# mole ratio, which means that the resulting solution will contain

#n_("HCl") = "0 moles HCl" -># completely consumed

#n_("NH"_3) = "0.0100 moles" - "0.000300 moles"#

#= "0.0097 moles NH"_3#

#n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"#

#="0.0103 moles NH"_4^(+)#

The total volume of the buffer will be

#V_"total" = "100.0 mL" + "3.00 mL" = "103.0 mL"#

The new concentrations of ammonia and ammonium cations will be

#["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"#

#["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"#

Notice that the concentration of ammonium cations remained virtually unchanged because the increase in the number of moles was counteracted by the increase in volume.

Now, you can find the change in pH by using the Henderson - Hasselbalch equation, which for a buffer that contains a weak base and its conjugate acid looks like this

#color(blue)(|bar(ul(color(white)(a/a)"pOH" = "p"K_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#

Notice that the the initial buffer solution had equal concentrations of weak base and conjugate acid. This means that its pOH was equal to

#"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))#

#"pOH" = "p"K_b#

After the strong acid is added to the buffer, the pOH of the buffer will be

#"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))#

#"pOH" = "p"K_b + 0.026#

You know that an aqueous solution at room temperature has

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#

For the initial solution, you have

#"pH"_ "initial" = 14 - "p"K_b#

For the final solution, you have

#"pH"_ "final" = 14 - ("p"K_b + 0.026)#

#"pH"_ "final" = 14 - "p"K_b - 0.026#

Therefore, you can say that the change in pH is equal to

#Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))#

#color(green)(|bar(ul(color(white)(a/a)color(black)(Delta_"pH" = -0.026)color(white)(a/a)|)))#

As predicted, the change in pH is negative because the pH of the buffer decreased as a result of the addition of a strong acid.

AHA! Actually, right after I wrote this question out, I figured it out.


The logic behind this is to express the change in pressure with respect to temperature at a constant volume in terms of #alpha# and #kappa#, since:

  • The container is rigid and full of liquid water. i.e. constant volume.
  • The water changes pressure due to the change in temperature. i.e. #(dP)/(dT) = ???#

Starting from the total derivative of the differential volume

#dV = ((delV)/(delT))_P dT + ((delV)/(delP))_T dP#,

we can divide by the partial differential temperature at a constant volume, #delT_V#, to find #((delP)/(delT))_V#.

#cancel(((delV)/(delT))_V)^(0) = ((delV)/(delT))_Pcancel(((delT)/(delT))_V)^(1) + ((delV)/(delP))_T stackrel("goal")overbrace(((delP)/(delT))_V)#

The term that cancels to #0# is because the change in volume must be #0# at a constant volume, and the term that cancels to #1# cancels because #(dT)/(dT) = 1#, and #1# at a constant volume is still #1#.

Thus, our big equation becomes:

#((delP)/(delT))_V = (-((delV)/(delT))_P)/((delV)/(delP))_T#

If we note that the definitions of #alpha# and #kappa# say #alpha = 1/V((delV)/(delT))_P# and #kappa = -1/V((delV)/(delP))_T#, then:

#((delP)/(delT))_V = (cancel(-V)alpha)/(cancel(-V)kappa) = alpha/kappa#

Now, we can do this trick where we multiply out a partial differential temperature at a constant volume to turn it into an exact differential, #dT_V#, and then integrate.

#dP_V = alpha/kappa dT_V#

#int_(P_1)^(P_2) dP_V = alpha/kappa int_(T_1)^(T_2) dT_V#

#P_2 - P_1 = alpha/kappa (T_2 - T_1)#

So the final expression is:

#bb(P_2 = P_1 + alpha/kappa (T_2 - T_1))#

And if we use this to evaluate the final pressure:

#color(blue)(P_2) = "1 atm" + (1.7xx10^(-4) "K"^(-1))/(4.7xx10^(-5) "atm"^(-1))("6 K")#

#= 22.7 ~~ color(blue)("23 atm")#

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