Question

# Please show all work so that i can understand. thank you A.) Which of the following...

Please show all work so that i can understand. thank you

A.) Which of the following processes have a ΔS > 0?

 a.CH3OH(l) → CH3OH(s) b.Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s) c. CH4(g) + H2O(g) → CO(g) + 3 H2(g) d. N2(g) + 3 H2(g) → 2 NH3(g) e. All of the above processes have a ΔS > 0.

B.) If ΔG > 0, which of the following statements are true?

 a. The reaction is at equilibrium. b. The reaction is spontaneous. c. The reaction is not spontaneous. d. Not enough information is given

C.) A reaction has ΔHrxn=−254kJ and ΔSrxn= 201J/K.

Calculate ΔGrxn at 41 ∘C.

 a. −8500kJ b. −262kJ c. −6.34×104kJ d. −317kJ

D.) Determine the equilibrium constant for the following reaction at 549 K.

CH2O(g) + 2 H2(g) → CH4(g) + H2O(g) ΔH° = -94.9 kJ; ΔS°= -224.2 J/K

 a. 1.07 x 109 b. 2.08 x 10-3 c. 481 d. 1.94 x 10-12 e. 9.35 x 10-10

A )

answer : c) CH4(g) + H2O(g) → CO(g) + 3 H2(g)

reason : moles of reactant = 2 , moles of products = 4 . so moles of products > moles of reactant entropy increases

B)

c) The reaction is not spontaneous.

C)

ΔH∘rxn = −254 kJ and ΔS∘rxn = 201J/K.

T = 41 + 273 = 314 K

ΔG∘rxn = ΔH∘rxn - T ΔS∘rxn

= - 254 - 314 x 201 x 10^-3

= - 317.1 kJ

ΔG∘rxn = - 317.1 kJ

D)

answer : b) 2.08 x 10^-3

ΔG∘rxn = -94.9 - 549 x (-224.2 x 10^-3)

= 28.19 kJ / mol

ΔG∘rxn = - R T ln Keq

28.19 = 8.314 x 10^-3 x 549 x ln Keq

Keq = 2.08 x 10^-3