Show all work for calculations

Question

2 AB (g) + B2 (g) → 2AB2 (g); H° = +492.0 kJ/mol, S° = +236.9 J/Kmol

The molar mass for AB for part (a) is 149.88

Answer the following questions regarding the reaction represented above.

(a) Calculate the quantity of heat released when 83.9 g of AB (g) is converted to AB2 (g).

(b) For the reaction at 25°C, the value of the standard free-energy change, G°, is +421.4 kJ/mol.

(i) Calculate the value of the equilibrium constant, K, for the reaction at 25°C. 

(ii) Indicate whether the value of G° would become more positive, less positive, or remain unchanged as the temperature is increased. Justify your answer. 

Answer 1

Short answers first: (a) 138 kJ, and by the data you were given it is absorbed, not released. (b)(i) K = 1.5 x 10^(-74). (b)(ii) less positive.

(a) Heat for 83.9 g of AB

n(AB) = 83.9 g / 149.88 g/mol = 0.5598 mol
The thermochemical equation is written for 2 mol of AB, so
   dH per mole of AB = +492.0 kJ / 2 = +246.0 kJ per mol AB
q = (0.5598 mol)(246.0 kJ/mol) = 137.7 kJ  ->  138 kJ (3 s.f.)

About the sign, because this is the part that costs points. dH° is +492.0 kJ/mol, so this reaction is endothermic and 138 kJ is taken in. The word "released" in the prompt does not agree with the data printed above it. Write the magnitude, state the direction, and say in one line why: "dH° is positive, so 138 kJ is absorbed." A grader gives you that. If your printed table actually reads dH° = -492.0 kJ/mol, the identical arithmetic gives 138 kJ released - the number does not change, only the sign does.

Note the two places students lose it here: dividing by 2 (the enthalpy is quoted per two moles of AB), and using the molar mass of AB rather than AB2, since AB is what you were handed 83.9 g of.

(b)(i) The equilibrium constant at 25 °C

dG = -RT ln K        ->        K = exp(-dG / RT)
dG = +421.4 kJ/mol = 421,400 J/mol
RT = (8.314 J/K mol)(298.15 K) = 2,478.8 J/mol
-dG / RT = -421,400 / 2,478.8 = -170.0
log10 K = -170.0 / 2.3026 = -73.83
K = 10^(-73.83) = 1.5 x 10^(-74)

K is astronomically small, which is exactly what a large positive dG° has to mean. At equilibrium this mixture is essentially pure reactants; the product is present at a level no instrument would ever see.

Check it for free. You were given dH°, dS° and dG°, which is one more number than you need - so use the spare one as a proof:

dG = dH - T dS
   = 492.0 kJ/mol - (298.15 K)(0.2369 kJ/K mol)
   = 492.0 - 70.6
   = +421.4 kJ/mol      matches the value the problem handed you

That agreement is your evidence you read the table correctly. It also forces the unit conversion that sinks more people on this problem than any other single step: dS° is in joules and dH° is in kilojoules. Divide dS° by 1000 before you subtract, every time.

One honest caveat about precision: with an exponent this large, using 298 K instead of 298.15 K shifts K from 1.5 x 10^(-74) to 1.4 x 10^(-74). Do not chase it. Report two significant figures and move on - the chemistry is identical.

(b)(ii) As temperature rises, dG° becomes LESS POSITIVE.

dG = dH - T dS,   with dS = +236.9 J/K mol  (positive)
Raise T  ->  the term (-T dS) grows more negative
         ->  dG drops, i.e. it becomes LESS POSITIVE

Both dH° and dS° are positive here, so this is the classic entropy-driven case: forbidden when it is cold, allowed once it is hot enough. You can even name the crossover, where dG° passes through zero:

T = dH / dS = 492,000 J/mol / 236.9 J/K mol = 2,077 K  (about 1,804 °C)

Below roughly 2,077 K the forward reaction is not spontaneous under standard conditions; above it, it is. Worth saying out loud in your justification - it turns a vague "less positive" into a specific claim.

The habit to carry out of this problem. You never need to compute anything to answer a "what does raising T do" question. dG° = dH° - T dS° is a straight line in T whose slope is -dS°, so the sign of dS° alone decides the direction: dS° positive means rising T always pushes dG° down, dS° negative always pushes it up, and if dS° were zero, T would not matter at all. Pair that with the sign of dH° and all four combinations fall out - spontaneous at every T, never spontaneous, only when hot (this one), only when cold. Learn the four cases and this question type takes ten seconds instead of ten minutes.