Question

For the reaction, heats of formation of $N_2{H}_4,H_2{O}_2,H_{2}O$ are 12, -45 and -57.8 $kcalmo{l}^{-1}$. Internal energy change for this reaction at $298K$ is:
$N_2{H}_4\left(l\right)+2H_2{O}_2\left(l\right)\to N_2\left(g\right)+4H_{2}O\left(g\right)$

• A. - 153.2 $kcalmo{l}^{-1}$
• B. - 641.142 $kJmo{l}^{-1}$
• C. -24.8 $kcalmo{l}^{-1}$
• D. -309 $kcalmo{l}^{-1}$

Answer 1

The correct option is A - 153.2 $kcalmo{l}^{-1}$

$N_2{H}_4\left(l\right)+2H_2{O}_2\left(l\right)⟶N_2\left(g\right)+4H_{2}O\left(g\right)\Delta H_{reaction}=\sum \Delta H_f\left(products\right)-\sum \Delta H_f\left(reactants\right)=\Delta H_f\left(N_2\right(g\left)\right)+4\Delta H_f\left(H_{2}O\right(g\left)\right)-\left[\Delta H_f\right(N_2{H}_4\left(l\right))+2\Delta H_f(H_2{O}_2\left(l\right)\left)\right]=0+4\times (-57.8)-2\times (-45)-12=-153.2kcal{mol}^{-1}Now,\Delta H=\Delta U+\Delta n_{g}RT\therefore \Delta U_{reaction}=\Delta H_{reaction}-\Delta n_{g}RT=(-153.2)-2\times 2\times 298\times {10}^{-3}=-153.2kcal{mol}^{-1}$