The reaction of cyanamide,$NH_2CN_{(s)}$,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 the enthalpy change for the reaction at $298 \ K$.
$NH_2CN_{(s)} + \frac{3}{2}O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$NH_2CN_{(s)} + \frac{3}{2}O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
The enthalpy change for a reaction $(\Delta H)$ is given by the expression:
$\Delta H = \Delta U + \Delta n_g RT$
Where:
$\Delta U = -742.7 \ kJ \ mol^{-1}$ (change in internal energy)
$\Delta n_g = \Sigma n_g(\text{products}) - \Sigma n_g(\text{reactants})$
For the reaction: $NH_2CN_{(s)} + 1.5 \ O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$\Delta n_g = (1 + 1) - 1.5 = 0.5 \ mol$
$T = 298 \ K$
$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$
Substituting the values:
$\Delta H = -742.7 + (0.5 \times 8.314 \times 10^{-3} \times 298)$
$\Delta H = -742.7 + 1.239$
$\Delta H = -741.46 \ kJ \ mol^{-1}$
$\Delta H = \Delta U + \Delta n_g RT$
Where:
$\Delta U = -742.7 \ kJ \ mol^{-1}$ (change in internal energy)
$\Delta n_g = \Sigma n_g(\text{products}) - \Sigma n_g(\text{reactants})$
For the reaction: $NH_2CN_{(s)} + 1.5 \ O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$\Delta n_g = (1 + 1) - 1.5 = 0.5 \ mol$
$T = 298 \ K$
$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$
Substituting the values:
$\Delta H = -742.7 + (0.5 \times 8.314 \times 10^{-3} \times 298)$
$\Delta H = -742.7 + 1.239$
$\Delta H = -741.46 \ kJ \ mol^{-1}$
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