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 ${27}^o$C. Assuming $R=2cal{K}^{-1}mo{l}^{-1}$, the value of $\Delta H$ for the above reaction will be:

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

Answer 1

The correct option is B 15.8 kcal

$A\left(g\right)+3B\left(g\right)⟶3C\left(g\right)+3D\left(g\right)$

$△E=17Kcal$ $T={27}^{o}C=300K$

$△H=?$ $R=2cal{K}^{-1}mo{l}^{-1}$

$△E=△H+nR△T$ $n$=no. of product-no. of reactant

$\Rightarrow 17Kcal=△H+\frac{2\times 2\times 300}{1000}Kcal$ $n=6-4=2$

$\Rightarrow 17Kcal=△H+1.2Kcal$

$△H=15.8Kcal$ .