Question

For the reaction,
$A\left(g\right)+2B\left(g\right)\to 2C\left(g\right)+3D\left(g\right)$
The value of $\Delta H$ at $27{}^{\circ }C$ is $19.0kcal$. The value of $\Delta E$ for the reaction would be

• A. $20.8kcal$
• B. $19.8kcal$
• C. $18.8kcal$
• D. $17.8kcal$

Answer 1

The correct option is D $17.8kcal$

Given reaction is,
$A\left(g\right)+2B\left(g\right)\to 2C\left(g\right)+3D\left(g\right)$
$\therefore \Delta n\left(g\right)=5-3=2$
We know,
$\Delta H=\Delta E+\Delta n_{g}RT\Rightarrow \Delta E=\Delta H-\Delta n_{g}RT=19-(2\times 2\times {10}^{-3}\times 300)=19-1.2kcal=17.8kcal.$