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 $27°C$ is $19.0\, kcal$. The value of $\Delta E$ for the reaction would be $\left(R = 2.0\, cal\, K^{-1}mol^{-1}\right)$

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

Answer 1

Answer: (D)

$\Delta H=\Delta E+\Delta n_{g}RT$
$19=\Delta E+\left(\frac{2\times2\times300}{1000}\right)kcal$
$19=\Delta E+1.2$
$\Delta E=19-1.2 kcal =17.8 \,kcal$