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 ${27}^{o}C$ is $19.0kcal$. The value of $\Delta E$ for the reaction would be: (Given $R=20cal{K}^{-1}$ ${mol}^{-1}$)

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

Answer 1

The correct option is D $17.8kcal$

The change in number of moles of gaseous species $\Delta n=\Sigma n_{\text{products}}-\Sigma n_{\text{reactants}}$

$\Delta n=2+3-(1+2)=\text{2 mol}$

$\Delta H=\Delta E+\Delta n\cdot RT$

$\Delta E=\Delta H-\Delta n\cdot RT$

$\Delta E=\text{19.0 kcal}-[\text{2 mol}\times \text{0.002 kcal/mol/K}\times \text{300 K}]$

$\Delta E=\text{17.8 kcal}$

Note:

$R=\text{2 cal/mol/K}\times \frac{\text{1 kcal}}{\text{1000 cal}}=\text{0.002 kcal/mol/K}$

$T=27{}^{o}C=(27+273)\text{K}=\text{300 K}$