Question

For the reaction,
`2A(g)+B(g) rarr 2D(g)`
`DeltaU^(Θ)=-10.5 kJ` and `DeltaS^(Θ)=-44.1 JK^(-1)`
Calculate `DeltaG^(Θ)` for the reaction, and predict whether the reaction may occur spontaneously.

Answer 1

Correct Answer - `0.164 kJ,` the reaction is not spontaneous.
For the given reaction ,
`2A_((g))+B_((g))rarr2D_((g))`
`Deltan_(g)=2-(3)`
1-mole
Substituting the value of `Delta^(theta)` in the expression of `DeltaH:`
`DeltaH^(theta)=DeltaU^(theta)+Deltan_(g)RT`
`=(-10.5 kJ)-(-1)(8.314xx10^(-3)kJ K^(-1)"mol"^(-1))(298 K)`
`=-10.5kJ-2.48 kJ`
`DeltaH^(theta)=-12.98 kJ`
Subtituting the value of `DeltaH^(theta) and DeltaS^(theta)` in the expression of `DeltaG^(theta):`
`DeltaG^(theta)=DeltaH^(theta)-TDeltaS^(theta)`
`=-12.98 kJ -(298 K)(-44.1 j K^(-1))`
`=-12.98 kJ +13.14 kJ`
`DeltaG^(theta)=+0.16 kJ`
Since `DeltaG^(theta)` for the reaction is positive, the reaction will not occur spontaneously.