Question

The reaction of cyanamide ${\mathrm{NH}}_2\mathrm{CN}\left(\mathrm{s}\right)$ with dioxygen was carried out in a bomb calorimeter and $△\mathrm{U}$ was found to be $-742.7{\mathrm{kJmol}}^{-1}$ at $298\mathrm{K}.$ Calculate enthalpy change for the reaction at ${\mathrm{NH}}_2\mathrm{CN}\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$.

Answer 1

Step 1: Enthalpy change:

• “Enthalpy change is the amount of heat evolved or absorbed in a reaction carried out at constant pressure.”

Step 2: Formula used for the calculation of enthalpy change:

• The enthalpy change for the reaction is expressed as,
• $\mathrm{\Delta H}=\mathrm{\Delta U}+{\mathrm{\Delta n}}_{\mathrm{g}}\mathrm{RT}$ , where $∆\mathrm{H}$ is the change in enthalpy, $∆\mathrm{U}$ is the change in internal energy, $\mathrm{R}$ is the gas constant, $∆{\mathrm{n}}_{\mathrm{g}}$ is the change in the number of moles, and $\mathrm{T}$ is the temperature.

Step 3: Given information:

The change in the number of moles$\left({\mathrm{\Delta n}}_{\mathrm{g}}\mathrm{RT}\right)$ can be given as,

$$\begin{gathered} \left.{\mathrm{\Delta n}}_{\mathrm{g}}=\sum {\mathrm{n}}_{\mathrm{g}}\text{(products}\right)-\sum {\mathrm{n}}_{\mathrm{g}}\text{(reactants)} \\ =(2-1.5)\mathrm{moles} \\ {\mathrm{\Delta n}}_{\mathrm{g}}=0.5\mathrm{moles} \\ \mathrm{\Delta U}=-742.7\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \mathrm{T}=298\hspace{0.17em}\mathrm{K} \\ \mathrm{R}=8.314\times {10}^{-3}\mathrm{kJ}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \end{gathered}$$

Step 4: Calculation of enthalpy change for the reaction:

$$\begin{gathered} \mathrm{\Delta H}=\left(-742.7\mathrm{kJ}{\mathrm{mol}}^{-1}\right)+(0.5\mathrm{mol})(298\mathrm{K})\left(8.314\times {10}^{-3}\mathrm{kJ}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right) \\ =-742.7+1.2 \\ \mathrm{\Delta H}=-741.5\mathrm{kJ}{\mathrm{mol}}^{-1} \end{gathered}$$

Therefore, the enthalpy change$(△\mathrm{H})$ for the reaction is $-741.5\mathrm{kJ}{\mathrm{mol}}^{-1}$.