Question

Calculate $\Delta H$for the reaction

$N_2{H}_4\left(l\right)+O_2\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(l\right)$

given the following data:

Equation	H(kJ)
$2N{H}_3\left(g\right)+3N_{2}O\left(g\right)\to 4N_2\left(g\right)+3H_{2}O\left(l\right)$	-1010
$N_{2}O\left(g\right)+3H_2\left(g\right)\to N_2{H}_4\left(l\right)+H_{2}O\left(l\right)$	-317
$2N{H}_3\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to N_2{H}_4\left(l\right)+H_{2}O\left(l\right)$	-143
$H_2\left(g\right)+\frac{1}{2}{O}_2\to H_{2}O\left(l\right)$	-286

Short answer

The for the reaction is -622.5 kJ.

Step-by-step solution

Simplification of the given reactions

Equation 1 multiplied by ¼. We get,

$\frac{1}{2}N{H}_3\left(g\right)+\frac{3}{4}{N}_{2}O\left(g\right)\to N_2\left(g\right)+\frac{3}{4}{H}_{2}O\left(l\right)\Delta H_1=-252.5kJ$

Equation 2 is reversed and multiplied by 3/4. We get,

$\frac{3}{4}{N}_2{H}_4\left(l\right)+\frac{3}{4}{H}_{2}O\left(l\right)\to \frac{3}{4}{N}_{2}O\left(g\right)+\frac{9}{4}{H}_2\left(g\right)\Delta H_2=237.75kJ$

Equation 3 is reversed and multiplied by ¼. We get,

$\frac{1}{4}{N}_2{H}_4\left(l\right)+\frac{1}{4}{H}_{2}O\left(l\right)\to \frac{1}{2}N{H}_3\left(g\right)+\frac{1}{8}{O}_2\left(g\right)\Delta H_3=35.75kJ$

Equation 4 is multiplied by 9/4. We get,

$\frac{9}{4}{H}_2\left(g\right)+\frac{9}{8}{O}_2\to \frac{9}{4}{H}_{2}O\left(l\right)\Delta H_4=-643.5kJ$

Determination of  ΔH

$\begin{array}{l}\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4\\ \Delta H=(-252.5kJ)+\left(237.75kJ\right)+\left(35.75kJ\right)+(-643.5kJ)\\ \Delta H=-622.5kJ\end{array}$

Thus, the for the reaction is -622.5 kJ