Question

For the reaction at 298 K,
2A(g) + B(g) → 2D(g)
ΔU = -10.5 kJ mol–1
and ΔS = -44.1J K–1 mol–1
Calculate standard gibbs energy for the reaction and predict whether the reaction is
spontaneous or not

#fast plz

Answer 1

Given : 2A(g) + B(g) => 2D(g)

• ΔU = -10.5 kJ mol–1
• ΔS = -44.1J K–1 mol–1
• T = 298 K

To find :

1. ∆G
2. Spontaneous or non spontaneous

Formula used :

• ∆G = ∆H - T∆S
• ∆H = ∆U + RT(∆n)

Solution :

∆n = number of mole of gaseous product - number of mole of gaseous reactant

∆n = 2 - 3 = (-1)

1) First of all we hve to find ∆H

∆H = ∆U + RT (∆n)

∆H = (-10.5kJ) + (8.314×298 × (2-3)

∆H = (-10500) + 8.314×298×(-1)

∆H = (-10500) - 2477.57

∆H = -12977.57 J/mol

Put value of ∆H in , ∆G = ∆H - T∆S

∆G = (-12977.57) - 298(-44.1)

∆G = -12977.57 + 13141.8

∆G = 164.23 J/mol

2) Non spontaneous , we can say this by 2 point

• ∆S is negative, which means non- spontaneous
• ∆G is positive , which means non- spontaneous

ANSWER :

1. ∆G = 164.23 J/mol
2. Non spontaneous