Question

For a gaseous reaction,
$A\left(g\right)+3B\left(g\right)\to 3C\left(g\right)+3D\left(g\right)$. $\Delta E$ is 17 Kcal at ${27}^{0}C$ assuming $R=2cal{K}^{-1}mo{l}^{-}$. The value of $\Delta H$ for the above reaction is :

• A. 15.8 Kcal
• B. 18.2 Kcal
• C. 20.0 Kcal
• D. 16.4 Kcal

Answer 1

The correct option is B 18.2 Kcal

Solution:- (B) $18.2\text{kcal}$

Given that:-

$\Delta E=17\text{kcal}$

$T=27℃=(273+27)=300K$

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

$A_{\left(g\right)}+3B_{\left(g\right)}⟶3C_{\left(g\right)}+3D_{\left(g\right)}$

$\therefore \Delta n_g=n_P-n_R=(3+3)-(1+3)=2$

As we know that,

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

$\therefore \Delta H=17+(2\times 300\times 2\times {10}^{-3})=18.2\text{kcal}$

Hence the value of $\Delta H$ for the given reaction will be $18.2\text{kcal}$.