Question

A) It is determined that for a particular process, ΔH = +185 kJ and ΔS = +1.80 J/K. At what temperature (in kelvin) does the reaction become spontaneous?

B)Which of the following statements are true for the reaction I2(g) → 2 I(g)
I. ΔH is positive
II. ΔH is negative
III. ΔS is positive
IV. ΔS is negative
V. The reaction is spontaneous at any temperature

C)

For which of the following processes would ΔS° be expected to be most positive

I) Na+(aq) + Cl-(aq) → NaCl(s)

II) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H20(g)

II) Cl2(g) + H2(g) → 2 HCl(g)

III) CO2(g) → CO2(s)

IV) H+(aq) + OH-(aq) → H2O(l)

Please explaine why . thank you!

Answer 1

A)

In order to compare equilibirum vs. spontaneous/nonspontaneous reactions, we better use a criteria.

Recall that if dSuniverse > 0, this is spontaneous, if dSuniverse = 0, this is inequilbirium and if dSuniverse < 0 this is never possible.

Then, recall that

dSuniverse = dSsurroundings + dSsystem

dSsystem = Sproducts - Sreactants

dSsurroundings = Qsurroundings/T = -dHsystem/T

therefore

dSystem =  -dHsystem/T + dSsystem

If we multiply by -T

dGrxn = dHrxn - T*dSrxn

Now, analysis of dG value... which is the "free energy" available for a process to follow

if dG <0 , this will be spontaneous

if dG = 0 , this is in equilibrium

if dG > 0, this will not be spontaneous

Now...

dG = dH - T*dS

Possible values are, dH = +/- and dS = +/-; T is always positive ( absolute value)

Analysis of cases:

Case 1.

if dH is positive (-) and dS is positive (+) --> this favours always a negative value of dG; spontaneous

Case 2.

if dH is positive (+) and dS is positive (-) --> this favours always a positive value of dG; not spontaneous

Case 3.

if dH is positive (+) and dS is positive (+) --> dG = dH - T*dS --> analysis must be done

if T is very low... then dH > T*dS; then this will be Positive value in dG; i.e. not spontaneous

if T is very high... then  dH < T*dS; then this will be Negative value in dG; i.e. spontaneous

Case 4.

if dH is positive (-) and dS is positive (-) --> dG = dH - T*dS --> analysis must be done

if T is very low... then dH > T*dS; then this will be Negative value in dG; i.e. spontaneous

if T is very high... then  dH < T*dS; then this will be Positive value in dG; i.e. non spontaneous

this is CASE 3.

ΔH = +185 kJ and ΔS = +1.80 J/K.

dG = dH - T*dS

dH - T*dS < 0

185*1000 - T*1.8 < T

185*1000 /1.8 < T

T > 102777.7 K

B)Which of the following statements are true for the reaction I2(g) → 2 I(g)

1 mol of gas (I2) form 2 mol of gas (I)

then, chaos incrases, dS =positive

this will break bonds between I-I, then this is exothermic, releases heat, dH is negative

choose:

II and III as correct answer

C)

Entropy is a measure of chaos; therefore, the more chaotic it becomes, the more entropy increases.

In general:

Entropy increases with the number of particles/moles in the system:

1 mol of AB has less entropy than 1 mol of A and 1 mol of B; due to higher amount of microstates

Entropy increases as the volume of particles increases. This can be compared with physical states

Volume of Solid < Volume of Liquid << Volume of Gas

Therefore, Gases have higher entropy than liquids and solids. Liquids have higher entropy than solids.

As T increases, the entropy increases, by definition, since it is a state function which depends on Temperature.

Mixture typically increases Entropy, so ordered states get higher amount of microstates.

Example could be Gas A + Gas B separated by a membrane, then eventually mixing A + B, microstates increases, meaning that entropy increases

Then...

I) Na+(aq) + Cl-(aq) → NaCl(s) NEGATIVE dS

II) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H20(g) 6 mol of gas form 12 mol of gas, good answer

II) Cl2(g) + H2(g) → 2 HCl(g) 2 mol of gas = 2 mol of gas, entorpy is almost 0

III) CO2(g) → CO2(s) -> chaos decreases, gas to solid

IV) H+(aq) + OH-(aq) → H2O(l) this is reversed, 2 mol of aqueos ions create 1 mol of liquid

best answer was II

II) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H20(g) 6 mol of gas form 12 mol of gas