The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter,
Question:

The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}_{(s)}$, with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7 \mathrm{~kJ}$ mol $^{-1}$ at $298 \mathrm{~K}$. Calculate enthalpy change for the reaction at $298 \mathrm{~K}$.

$\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}$

 

Solution:

Enthalpy change for a reaction $(\Delta H)$ is given by the expression,

$\Delta H=\Delta U+\Delta n_{g} R T$

Where,

$\Delta U=$ change in internal energy

$\Delta n_{g}=$ change in number of moles

For the given reaction,

$\Delta n_{g}=\sum n_{g}$ (products) $-\sum n_{g}$ (reactants)

= (2 – 1.5) moles

$\Delta n_{g}=0.5$ moles

And,

$\Delta U=-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

T = 298 K

$\mathrm{R}=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Substituting the values in the expression of $\Delta H$ :

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

$=-742.7+1.2$

$\Delta H=-741.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

 

 

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