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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics5 Marks
Question
The reaction of cyanamide, $NH _2 CN ( s )$, with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7 kJ mol ^{-1}$ at 298 K . Calculate enthalpy change for the reaction at 298 K .
$NH_2 CN(g)+\frac{3}{2} O_2(g) \longrightarrow N_2(g)+CO_2(g)+H_2 O(l)$
✓

Answer

Enthalpy change for a reaction $(\Delta\text{H}) $ is given by the expression,$\Delta\text{H} = \Delta\text{U} + \Delta\text{n}_\text{g}\text{RT}$
Where,
$\Delta\text{U} =$ change in internal energy
$\Delta\text{n}_\text{g} =$ change in number of moles
For the given reaction,
$\Delta\text{n}_\text{g} = \sum\text{n}_\text{g} (\text{products}) – \sum\text{n}_\text{g} (\text{reactants})$
$= (2 – 1.5) \ \text{moles}$
$\Delta\text{n}_\text{g} =0.5 \ \text{moles}$
And,
$\Delta\text{U} = \ –742.7\text{kJ} \ \text{mol}^{–1}$
$\text{T} = 298\text{K}$
$\text{R} = 8.314 × 10^{–3}\text{kJ} \ \text{mol}^{–1} \ \text{K}^{–1}$
Substituting the values in the expression of $\Delta\text{H}:$
$\Delta\text{H} = (–742.7\text{kJ} \ \text{mol}^{–1}) + (0.5 \ \text{mol}) (298\text{K})\$8.314 × 10^{–3}\text{kJ} \ \text{mol}^{–1} \ \text{K}^{–1})$
$= –742.7 + 1.2$
$\Delta\text{H} = –741.5\text{kJ} \ \text{mol}^{–1}$

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