Question

For the reaction, $A\left(g\right)+2B\left(g\right)\to 2C\left(g\right)+3D\left(g\right)$ the change of enthalpy at ${27}^o$C is $19$ kcal. The value of $\Delta$E is:

• A. $21.2$ kcal
• B. $17.8$ kcal
• C. $18.4$ kcal
• D. $20.6$ kcal

Answer 1

The correct option is B $17.8$ kcal

The reaction is, $A\left(g\right)+2B\left(g\right)\to 2C\left(g\right)+3D\left(g\right)$

Here, $\Delta n_g=n_p-n_r=5-3=2$

$\Delta H=19kcal=19\times {10}^{3}cal$

$\therefore$ The value of internal energy change,

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

$=19\times {10}^3-(2\times 2\times 300)$

$=17.8kcal$.