Question

$\Delta H^0$ for the raection $X_{\left(g\right)}+Y_{\left(g\right)}\rightleftharpoons Z_{\left(g\right)}$ is -4.6Kcal, the value of $\Delta U^0$ of the reaction at ${227}^{o}C$ is $R=2cal{mol}^{-1}{K}^{-1}$

• A. -3.6kcal
• B. -5.6kcal
• C. -4.6kcal
• D. -2.6kcal

Answer 1

The correct option is A -3.6kcal

Solution:- (A) $-3.6kcal$

$X_{\left(g\right)}+Y_{\left(g\right)}\rightleftharpoons Z_{\left(g\right)}$

$\Delta n_g=n_P-n_R=1-(1+1)=-1$

Given:-

$\Delta H=-4.6Kcal$

$R=2cal/mol-K=2\times {10}^{-3}kcal/mol-K$

$T=227℃=(273+227)=500K$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

$\Rightarrow \Delta U=-4.6-(-1\times 2\times {10}^{-3}\times 500)$

$\Rightarrow \Delta U=-4.6+1=-3.6kcal$

Hence the value of $|DeltaU$ for the given reaction is $-3.6kcal$.