Consider the following ionic reaction in aqueous medium under standard state MnO

Question

Consider the following ionic reaction in aqueous medium under standard state MnO_4^- + Mn^2+ + O_2 + 1.20 KJ/mol What is the entropy change of this redox reaction if the free energy change is 1.80 kj/mol? A. 0.01067 j/k B. - 0.00201 kj/k C. 10.06 j/k D. 0.00207 kj/k E. 10.06 kj/k. Consider the reaction C_5 H_s (g) + 5 O_2 (g) --- 3 CO_2 (g) + 4 H_2 O (i) + 1.80 kJ The entropy change for the reaction is - 12.0 j/k. At what temperature (s) would you expect the reaction to be spontaneous? A. T > 150 k C. T

Explanation / Answer

Solution:- (1) heat is on product side it means reaction is exothermic and so delta H = -1.20 kJ/mol

delta G = 1.80 kJ/mol

from Gibb's free energy equation...

delta G = delta H - T*delta S

at standard conditions, T = 298 K

1.80 kJ = (-1.20 kJ) - 298K(delta S)

1.80 kJ + 1.20 kJ = - 298K(delta S)

3.0 kJ = -298K(delta S)

delta S = - 3.0 kJ/298K = -0.01006 kJ/K or -10.06 J/K

choice C is correct.

(2) It's again based on the same equation, delta G = delta H - T*delta S

at equilibrium, delta G = 0

so, delta H = T*delta S

T = delta H/delta S

given, delta H = -1.80 kJ = -1800 J

delta S = -12.0 J/K

T = -1800J/-12.0 J/K

T = 150 K

so, for the reaction to be spontaneous delta G has to be less than 0. negative value of delta H and positive value of delta S favors negative value of delta G. Since, delta S is also negative here so the temperature has to be less than 150 K that means T<150 K

so, the right choice is D.

(3) to calculate the standard enthalpy change we could use the standard enthalpy of formation values from the table that would be given in the book or available online.

First of all we need to balance the equation.

C5H12(g) + 8O2(g) ----> 5CO2(g) + 6H2O(l)

delta Hrxn = sum of delta H of products - sum of delta H of reactants

delta Hrxn = [5(delta H of CO2) + 6(delta H of H2O)] - [(delta H of C5H12) + 8(delta H of O2)]

delta Hrxn = [5(-393.5 kJ) + 6(-285.8 kJ)] - [(-35.1 kJ) + 8(0)]

delta Hrxn = [(-1967.5 kJ) + (-1714.8 kJ)] - [-35.1 kJ]

delta Hrxn = -3682.3 kJ + 35.1 kJ = -3647.2 kJ

choice C is closer so that would be the right one.

(4) we know that, delta G0 = - RT ln k

where k is the equilibrium constant, T is kelvin temperature and R is the universal gas constant. Here the value of R is 8.314 x 10-3 kJ.mol-1.K-1.

T = 298 K and delta G0 = -3.10 kJ.mol-1

let's plug in the values in the formula..

-3.10 kJ.mol-1 = - 8.314 x 10-3 kJ.mol-1.K-1.298K ln k

-3.10 kJ.mol-1 = -2.478 kJ.mol-1 ln k

ln k = -3.10 kJ.mol-1/-2.478 kJ.mol-1

ln k = 1.251

k = e1.251

k = 3.49

choice E is correct.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.