Question

$\Delta H^0$ for the reaction $X_{\left(g\right)}+Y_{\left(g\right)}\rightleftharpoons Z\left(g\right)$ is $-4.6$ K.cal, the value of $\Delta U^0$ of the reaction at ${227}^0$C is:

$(R=2cal.mo{l}^{-1}{K}^{-1})$

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

Answer 1

The correct option is A $-3.6$ kcal

Given: $\Delta H^o=-4.6KCal=-4600cal$

$R=2Calmo{l}^{-1}{K}^{-1}$

$T={227}^{o}C=227+273K=500K$

Using the relation, $\Delta H^o=\Delta U+\Delta n_{g}RT$

$\Delta n_g=$ Number of gaseous moles

$X\left(g\right)+Y\left(g\right)\rightleftharpoons Z\left(g\right)$

$\Delta n_g=1-1-1=-1$

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

$=-4600cal-(-1\times 2calmo{l}^{-1}{K}^{-1}\times 500K)$

$=-4600cal+1000cal=-3600cal$

$\Delta U=-3.6Kcal$