Featured Answers

14
Active contributors today

Answer:

Here's my explanation

Explanation:

According to Ohm's law the current passed through a material is directly proportional to the voltage.

upload.wikimedia.org

You can see the mathematical equation that describes this proportionality

#I = V/R#

#V = IR#

Where #I# is the current in the units of "amperes"
Where #V# is voltage
and #R# is the resistance of the material in units of "ohms"

The other equation

#E = Jsigma#

Where E is electric field at the given location
J is current density
σ (sigma) is a material-dependent parameter called the conductivity

#R =\rho\frac l A #

R is the electrical resistance of a uniform specimen of the material
#l# is the length of the piece of material
A is the cross-sectional area of the specimen

Simplifying

#\rho = \R\frac A l #

The resistance of a given material increases with length, but decreases with increasing cross-sectional area

If the #A = 1m^2# and #l=1# the resistance of the conductor is equal to the the ρ

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at #20^@C#

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html

Note that ρ is different for conductors at different temperature

ρ increases as the temperature increases
ρ decreases as the temperature decreases

#sigma = 1/ρ#

Where #sigma# is conductivity
ρ is resistivity

There are many version of the Ohm's law

#P = (ΔV^2) / R#

#DeltaV# is the potential difference between the two points in the conductor

#P = I^2 • R#

Now lets come to the main problem.

#ΔV =( ΔPE )/ Q#

Where Q is charge
PE is energy

By this you can understand that Voltage is proportional to charge which is proportional to coulombs

As

#C = V * Q#
There is a relationship between V and C

As the V increases and the Q is thought to be constant the C increases. Means more voltage more coulombs

Now let me show you an example

www.docbrown.info

Suppose that #2C#/s is applied to a solution of #CuSO4# solution for 2 seconds.

The reaction

= #2H_2O + 2CuSO_4 rarr O_2 + 2H_2SO4 + 2Cu#

Net ionic reaction

#2H_2O + Cu^(2+) rarr 2H_2 + 2Cu#

Each mole of #Cu^(2+)# produces one mole of #Cu#

#2H_2O# is reduced to #H_2#

#H_2O + rarr H_2 + 2e^-#
#Cu^(2+) + 2e^-) rarr Cu #

Thus
#2H_2O rarr 2H_2 + 4e^-#
#2Cu^(2+) + 4e^-) rarr Cu #

The whole reaction

#2H_2O + 2Cu^(2+) + 4e^-) rarr 2H_2 + Cu + 4e^-)#
#2H_2O + 2Cu^(2+) rarr 2H_2 + Cu #

As 2C/s is applied for 2s

#(2C)/cancel(s) xx 2cancel(s)#

= #4C#

Use the Faraday's constant to calculate the moles of electrons lost and gained in total.

#( 4 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00004145077 mole of electrons"#

1 mole of #Cu^(2+)# is reduced per 2 mole electrons

Thus moles of #Cu# formed

#"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)#

#1/2 xx 0.00004145077 = "0.00002072538 mol of Cu"#

Now see what happens if we apply more charge

Consider the charge as 3C/s for 2s

Do the same calculations

As 3C/s is applied for 2s

#(3C)/cancel(s) xx 2cancel(s)#

= #6C#

#( 6 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00006217616 mole of electrons"#

#"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)=" 0.00003108808 mol of Cu"#

#"0.00003108808 mol of Cu" > "0.00002072538 mol of Cu"#

Thus more the current,voltage and coulomb the more the mol of copper formed thus more the mass of copper formmed

Plotting a graph

enter image source here

If my handwriting is too bad

y = mol of Cu formed from reduction of #Cu^(2+)#
x = mol of electrons used in the process

I will stick to diatomic molecules for simplicity. Some general steps are:

  1. Choose the set of atomic valence orbitals that each atom comes in with. Assume core orbitals don't interact. Determine your coordinate axes.
  2. Have an approximate idea of the relative orbital energies if working with a heteronuclear diatomic molecule.
  3. By conservation of orbitals, each of the two corresponding atomic orbital interacts to produce one bonding and one antibonding molecular orbital in the middle.
  4. By conservation of electrons, the valence electrons that are contributed by each atom equal the total number of valence electrons distributed in the MO diagram.
  5. Fill the diagram with electrons as usual in accordance with the Aufbau principle, Hund's rule, etc.

Some general rules or tips are:

  • Antibonding orbitals usually are approximately as higher in energy as bonding orbitals are lower in energy than the original atomic orbital energies.
  • Nonbonding orbitals have either a similar energy to the original atomic orbital energy, or have both bonding and antibonding contributions from the interacting atomic orbitals.
  • Large atomic orbital energy offsets correspond to relative molecular orbital energy offsets. This helps with orbital energy ordering.

Let's construct an MO diagram for #"CO"#.

CHOOSE YOUR ATOMIC ORBITALS & AXES

  • Carbon has #2s# and #2p# atomic valence orbitals.
  • Oxygen has #2s# and #2p# atomic valence orbitals.

ATOMIC ORBITAL ENERGY DIAGRAM

From the original atomic orbital energies, we then construct the two atomic orbital energy diagrams:

(On an exam, you may only be expected to work with homonuclear diatomic molecules, in which case you probably don't have to worry about relative energies.)

GENERATION OF MOLECULAR ORBITALS

Then, each pair of atomic orbitals generates molecular orbitals as follows. Here is an #ns# interaction:

For lithium through nitrogen (including nitrogen), orbital mixing occurs, so that the #sigma_(np_z)# orbital is higher in energy than the #pi_(np_x)# and #pi_(np_y)# orbitals.

Therefore, carbon has its energy ordering switched compared to oxygen, and so, the #np# interactions look like this:

OVERALL MO DIAGRAM

Now put it together to get:

Note that on an exam, you probably aren't expected or required to know exactly how the interactions go. It's enough to know the relative energies of the molecular orbitals compared to the atomic orbitals, and to know how many valence electrons go in.

DISCLAIMER: LONG ANSWER!

#"NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH"(aq)#

#a)# This part is just asking you to write out your starting equation. It turns out that the Nernst equation is also for half-cells.

The general Nernst equation relates the non-standard #E_"cell"# to the #E_"cell"^@#, i.e. at standard conditions of #"1 M"# concentrations, #"1 atm"# pressure, and #25^@ "C"#:

#E_"cell" = E_"cell"^@ - (RT)/(nF)lnQ#

We now know how to write #Q#, the reaction quotient:

#Q = (["NADH"])/(["NAD"^(+)]["H"^(+)])#

Hence, the Nernst equation for this problem, which is for a half-cell, is:

#bb(E_"red" = E_"red"^@ - (RT)/(nF)ln((["NADH"])/(["NAD"^(+)]["H"^(+)]))#

#b)# Given a #"Pt" | "NADH, NAD"^(+),"H"^(+)# half-cell at #"pH"# #7#, we know:

  • that the platinum electrode is inert and does not participate in the reaction.
  • that the concentration of #bb("H"^(+))# is #["H"^(+)] = 10^(-"pH") = 10^(-7) "M"#.
  • that from the half-cell reaction, the number of electrons transferred is TWO.

We are also given #E_"red"^@ = -"0.11 V"#, and that all other components are at #"1 M"# and the solutions are all at #"298 K"#.

So, we just use the equation we wrote down in part #(a)#:

#color(blue)(E_"red") = -"0.11 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M")/("1 M"cdot 10^(-7) "M"))#

#= color(blue)(-"0.32 V")#

This indicates that the reduction of #"NADH"# is nonspontaneous using these concentrations in these conditions.

#c)# We know that cytochrome c with #"Fe"^(3+)# was reduced by #"NADH"# to cytochrome c with #"Fe"^(2+)#.

Since the important reaction is the reduction of the iron center, let's focus on that half-reaction. However, don't forget our other half-reaction. Since #"NADH"# was the reducing agent, it gets oxidized.

Thus, we flip our first half reaction, which was a reduction to begin with, and turn it into an oxidation half-reaction.

#2("Fe"^(3+)(aq) + cancel(e^(-)) -> "Fe"^(2+)(aq))#
#"NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + cancel(2e^(-))#
#"-------------------------------------------------------------"#
#color(blue)(2"Fe"^(3+)(aq) + "NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + 2"Fe"^(2+)(aq))#

I'll leave you to check that the mass and charge are balanced.

#d)# Now, we are given the #"Pt" | "cyt c"("Fe"^(2+)), "cyt c"("Fe"^(3+))# half-cell, with an #E_"red"^@ = "0.25 V"# relative to the standard hydrogen electrode:

#"Fe"^(3+)(aq) + e^(-) -> "Fe"^(2+)(aq)#, #E_"red"^@ = "0.25 V"#

We now essentially repeat what we did in parts #(a,b)# with the reaction we wrote in part #(c)#.

This is a pretty bulky reaction quotient expression...

#Q = (["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"])#

Thus, the Nernst equation for this problem is:

#E_"cell" = E_"cell"^@ - (RT)/(nF)ln((["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"]))#

We first have to figure out #E_"cell"^@#. We were given in part #(b)# that #E_"red"^@ = -"0.11 V"# for the #"NAD"^(+) -> "NADH"# reduction, and in part #(d)# that #E_"red"^@ = +"0.25 V"# for the #"Fe"^(3+) -> "Fe"^(2+)# reduction.

Since we know from part #(c)# that #"NADH"# was oxidized, we can flip its reduction potential sign to obtain #E_"ox"^@ = +"0.11 V"#. Therefore, the standard cell potential is:

#E_"cell"^@ = E_"red"^@ + E_"ox"^@#

#= "0.25 V" + [-stackrel(E_"red"^@)overbrace((-"0.11 V"))]#

#=# #+"0.36 V"#

Now, we can calculate the non-standard cell potential at #"pH"# #7# and (presumably) #"298 K"# again. The same number of electrons were transferred (two!), and the concentration of #"H"^(+)# is #10^(-7) "M"# again.

#color(blue)(E_"cell") = "0.36 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M"cdot10^(-7)"M"cdot("0.004 M")^2)/(("0.01 M")^2cdot"1 M"))#

#= color(blue)(+"0.59 V")#

Thus, the reduction of the iron center in cytochrome c by #"NADH"# at these concentrations in these conditions is spontaneous.

Answer:

See below:

Explanation:

#sf(NAD^(+)+H^++2erightleftharpoonsNaDH" "E^@=-0.11color(white)(x)V)#

#sf((a))#

The potential E for a 1/2 cell is given by:

#sf(E=E^@-(RT)/(zF)ln([["reduced form"]]/(["oxidised form"]))#

At 298K this can be simplified to:

#sf(E=E^@-0.0591/(z)log([[NaDH]]/([NaD^+][H^+])))#

Where z is the no. of moles of electrons transferred which, in this case =2.

#sf((b))#

Putting in the numbers:

#sf(E=-0.11-0.591/2log((1)/((1)xx(10^(-7))))#

#sf(E=-0.11-0.2068=color(red)(-0.32color(white)(x)V))#

#sf((c))#

It helps to consider the #sf(E^@)# values:

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(NAD^(+)+H^++2erightleftharpoonsNaDH" "E^@=-0.11color(white)(x)V)#

#sf(cFe^(3+)+erightleftharpoonscFe^(2+)" "E^@=+0.25color(white)(x)color(white)(x)V)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

The most +ve 1/2 cell is the one that will take in the electrons so you can see that the 1st 1/2 cell will move right to left and the 2nd 1/2 cell will move left to right in accordance with the arrows.

The two 1/2 equations are therefore:

#sf(NaDHrarrNaD^++H^++2e)#

#sf(cFe^(3+)+erarrcFe^(2+))#

To get the electrons to balance we X the 2nd 1/2 cell by 2 then add:

#sf(NaDH+2cFe^(3+)+cancel(2e)rarrNaD^(+)+H^++cancel(2e)+2cFe^(2+))#

#sf((d))#

Now we have to use the full Nernst Equation:

#sf(E_(cell)=E_(cell)^@-(RT)/(zF)lnQ)#

This can be simplified at 298K to:

#sf(E_(cell)=E_(cell)^@-0.0591/(z)logQ)#

Where z = 2.

To find #sf(E_(cell)^@)# there are different conventions around but they all rely on finding the arithmetic difference between the two #sf(E^@)# values for the 1/2 cells.

The clearest way is to subtract the least +ve #sf(E^@)# value from the most +ve.

Using the values in (c):

#sf(E_(cell)^@=+0.25-(-0.11)=+0.36color(white)(x)V)#

From the complete equation the reaction quotient Q is written:

#sf(Q=([NaD][H^+][cFe^(2+)]^2)/([cFe^(3+)]^2[NaDH]))#

pH = 7 #:.##sf([H^+]=10^(-7)color(white)(x)"mol/l")#

Putting the numbers into The Nernst Equation:

#sf(E_(cell)=+0.36-0.0591/2log((1xx10^(-7)xx0.004^2)/(0.01^2xx1)))#

#sf(E_(cell)=+0.36+0.22885 color(white)(x)V)#

#sf(E_(cell)=color(red)(+0.59color(white)(x)V)#

Answer:

The basic formula is #color(red)("M"_2"O"#.

Explanation:

To answer this, we can take a look at the ionic charges of the elements involved.

Since oxygen is in group 16, it will have #color(red)(6# valence electrons:

upload.wikimedia.org

Recall that an atom will want to gain or lose electrons to fulfill the octet rule; it wants to do whatever is easiest to obtain #8# electrons in its outermost shell.

For an oxygen atom, the easiest way to get #8# electrons in the outer shell is to obtain #2# more electrons. This will in turn cause the #sfcolor(blue)("oxygen"# to have an ionic charge of #color(blue)(2-#, because there are #color(blue)(2# more electrons than protons now.

#-----#

For the alkali metals, which are located in group 1 of the periodic table, they each have #1# valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:

upload.wikimedia.org

In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol #color(red)("M"# for the alkali metal) to have an ionic charge of #color(red)(1+#, because there is #color(red)(1# less electron than the number of protons.

The formula:

The compound the alkali metal #color(red)("M"# and oxygen #color(blue)("O"# form must be electrically neutral. That is to say, the charges must balance out to #0#.

The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to put the charge of the anion (the negative ion, in this case is #"O"^(2-)#), #2-# as the subscript of the cation, and the charge of the cation (the positive ion, in this case #"M"^+#) as the subscript of the *anion*. Here's what I mean, using sodium (#"Na"#) as an example:

http://web.fscj.edu

#sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"#
#sfcolor(green)(" is M"_2"O"#.

The main postulates or assumptions are (for ideal gases):

  1. Gases are constantly in motion, colliding elastically in their container.
  2. They do not interact (i.e. no intermolecular forces are considered), and are assumed to be point masses with negligible volume compared to the size of their container.
  3. Ensembles of gases produce a distribution of speeds, and the average kinetic energy of this ensemble is proportional to the translational kinetic energy of the sample.

As for "one kinetic gas equation", there is not just one. But I can derive the Maxwell-Boltzmann distribution, and from there one could derive many other equations... some of them being:

  • collision frequency
  • average speed
  • RMS speed
  • most probable speed

The latter three are derived in more detail here.


The following derivation is adapted from Statistical Mechanics by Norman Davidson (1969). Fairly old, but I kinda like it.

Consider the classical Hamiltonian for the free particle in three dimensions:

#H = (p_x^2)/(2m) + (p_y^2)/(2m) + (p_z^2)/(2m)#,

where #p = mv# is the momentum and #m# is the mass of the particle.

The distribution function will be denoted as #(dN)/N#, where #N# is the total number of particles in the system, and #dN# is a differential function whose coordinates will be defined.

From Statistical Mechanics by Norman Davidson (pg. 150), we have that

#(dN)/N = (e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n)/zeta#,

where:

  • #beta = 1//k_BT# is a constant, with #k_B# as the Boltzmann constant and #T# as the temperature in #"K"#.
  • #H(p_1, . . . , p_n; q_1, . . . , q_n)# is the Hamiltonian for a system with #3N# momentum coordinates and #3N# position coordinates, containing #N# particles in 3 dimensions.
  • #zeta = int cdots int e^(-betaH(p_1, . . . , p_n; q_1, . . . , q_n)) dp_1 cdots dp_n dq_1 cdots dq_n# is the classical phase integral.

This seems like a lot, but fortunately we can look at three coordinates for simplicity.

A phase integral is an integral over #(-oo,oo)# in phase space, a coordinate system where two conjugate variables are orthogonal to each other. In this case, position #q# and momenta #p_q# are conjugate variables.

With #e^x -= "exp"(x)#, #dN# can then be defined as:

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(int_(-oo)^(oo) int_(-oo)^(oo) int_(-oo)^(oo) "exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)#

Now, let's take the bottom integral and separate it out:

#zeta = int_(-oo)^(oo) e^(-p_x^2//2mk_BT)dp_x cdot int_(-oo)^(oo) e^(-p_y^2//2mk_BT)dp_y cdot int_(-oo)^(oo) e^(-p_z^2//2mk_BT)dp_z#

In phase space, each of these integrals are identical, except for the particular notation. These all are of this form, which is tabulated:

#2int_(0)^(oo) e^(-alphax^2)dx = cancel(2 cdot 1/2) (pi/alpha)^(1//2)#

Notice how the answer has no #x#. Therefore, for the three integrals in #zeta#, we then let #alpha = 1//2mk_BT# so that...

#zeta = (2pimk_BT)^(3//2)#

So far, we then have:

#(dN(p_x,p_y,p_z; x,y,z))/(N) = ("exp"(-(p_x^2 + p_x^2 + p_z^2)/(2mk_BT))dp_xdp_ydp_z)/(2pimk_BT)^(3//2)#

Next, it is convenient to transform into velocity coordinates, so let #p_q = mv_q# so that...

#(dN(v_x,v_y,v_z; x,y,z))/(N) = (e^(-m^2(v_x^2 + v_y^2 + v_z^2)//2mk_BT)m^3dv_xdv_ydv_z)/(2pimk_BT)^(3//2)#

#= (m/(2pik_BT))^(3//2)e^(-m(v_x^2 + v_x^2 + v_z^2)//2k_BT)dv_xdv_ydv_z#

Now, we assume that the gases are isotropic, meaning that no direction is any different from any other.

To enforce that, suppose we enter velocity space, a coordinate system analogous to spherical coordinates, but with components of #v# being along the coordinate axes.

#v_x = vsintheta_vcosphi_v#
#v_y = vsintheta_vsinphi_v#
#v_z = vcostheta_v#

http://nptel.ac.in/

With the differential volume element being #v^2 dv sintheta_v d theta_v dphi_v#, this results in this godawful mess:

#(dN(v, theta_v, phi_v))/(N)#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v cos^2phi_v + v^2 sin^2theta_v sin^2phi_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

Fortunately, using #sin^2u + cos^2u = 1#, we obtain a cleaner representation:

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v(cos^2phi_v + sin^2phi_v) + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#= (m/(2pik_BT))^(3//2) cdot e^(-m(v^2sin^2theta_v + v^2 cos^2theta_v)//2k_BT)v^2 dv sintheta_v d theta_v d phi_v#

#=> ul((dN(v, theta_v, phi_v))/(N) = (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv sintheta_v d theta_v d phi_v)#

We're almost there! Now, the integral over allspace for #theta_v# is over #0 -> pi#, which gives

#int_(0)^(pi) sintheta_vd theta_v = 2#

And the integral over allspace for #phi_v# is

#int_(0)^(2pi) dphi_v = 2pi#

So, we integrate both sides to get:

#color(blue)(barul(|stackrel(" ")(" "(dN(v))/(N) -= f(v)dv = 4pi (m/(2pik_BT))^(3//2)e^(-mv^2//2k_BT)v^2 dv" ")|))#

And this is the Maxwell-Boltzmann distribution, the one you see here:

https://upload.wikimedia.org/

DERIVATION OF SOME OTHER EQUATIONS

This is only a summary.

Gas Speed Equations

#<< v >> = int_(0)^(oo) vf(v)dv = sqrt((8k_BT)/(pim)#

#v_(RMS) = << v^2 >>^(1//2) = (int_(0)^(oo) v^2f(v)dv)^(1//2) = sqrt((3k_BT)/m)#

#v_(mp)#: Take the derivative of #f(v)#, set the result equal to #0#, and solve for #v# to get

#v_(mp) = sqrt((2k_BT)/m)#

View more
Questions
Ask a question Filters
Loading...
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions