#### Question

For the reaction

2A_{(}_{g}_{)} + B_{(}_{g}_{)} → 2D_{(}_{g}_{)}

Δ*U*^{θ} = –10.5 kJ and Δ*S*^{θ}= –44.1 JK^{–1}.

Calculate Δ*G*^{θ} for the reaction, and predict whether the reaction may occur spontaneously.

#### Solution 1

For the given reaction,

2 A_{(}_{g}_{)} + B_{(}_{g}_{)} → 2D_{(}_{g}_{)}

Δ*n*_{g} = 2 – (3)

= –1 mole

Substituting the value of Δ*U*^{θ} in the expression of Δ*H*:

Δ*H*^{θ} = Δ*U*^{θ} + Δ*n*_{g}R*T*

= (–10.5 kJ) – (–1) (8.314 × 10^{–3} kJ K^{–1} mol^{–1}) (298 K)

= –10.5 kJ – 2.48 kJ

Δ*H*^{θ} = –12.98 kJ

Substituting the values of Δ*H*^{θ} and Δ*S*^{θ} in the expression of Δ*G*^{θ}:

Δ*G*^{θ} = Δ*H*^{θ} – *T*Δ*S*^{θ}

= –12.98 kJ – (298 K) (–44.1 J K^{–1})

= –12.98 kJ + 13.14 kJ

Δ*G*^{θ} = + 0.16 kJ

Since Δ*G*^{θ} for the reaction is positive, the reaction will not occur spontaneously.

#### Solution 2

`triangleH^(Theta) = triangleU^(Theta) + triangle^"ng" RT`

`triangleU^(Theta) = -10.5 kJ; triangle^"ng" = 2 - 3` = -1 mol

`R = 8.314 xx 10^(-3) kJ K^(-1) mol^(-1)`; T = 298 K

`:. triangle H^(Theta) = (-10.5 kJ) + [(- 1 mol) xx (8.314 xx 10^(-3) kJ K^(-1) mol^(-1))xx(298K)]`

= -10.5 kJ - 2.478 kJ = -12.978 kJ

According to Gibbs Helmhoitz equation:

`triangle G^(Theta) = triangleH^(Theta) - T triangle S^(Theta)`

`triangle G^(Theta) = (-12.978 kJ) - (298 K) xx (-0.0441 kJ K^(-1))`

= - 12.978 + 13.112 = - 12.978 + 13.142 = 0.164 kJ

Since `triangleG^(Theta)` is postive the reaction is non-spontaneous in nature