The reaction of cyanamide $N H_{2} C N (s)$ with dioxygen was carried out in a bomb calorimeter and $△ U$ was found to be $- 742.7 k J m o l^{- 1}$ at 298 K.
$N H_{2} C N (g) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$

Calculate enthalpy change for the reaction at 298 K?

Open in App

Solution

For given reaction, $Δ n = 1 + 1 - 1.5 = 0.5$ .

$Δ H = Δ U + Δ n_{g} R T$

$= - 742.7 + 0.5 \times 8.314 \times 10^{- 3} \times 298 = - 742.7 + 1.2 = - 741.5 k J m o l e^{- 1}$

Suggest Corrections

Similar questions

Q. The reaction of cyan-amide,

N H_{2} C N (s)

, with dioxygen was carried out in a bomb calorimeter, and

Δ U

was found to be

- 742.7 k J {m o l}^{- 1}

at 298 K. Calculate enthalpy change for the reaction at 298 K.

N H_{2} C N (g) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + {C O}_{2} (g) + H_{2} O (l)

The reaction of cyanamide, $N H_{2} C N_{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $Δ U$ was found to be $- 742.7 k J m o l^{- 1}$ at 298 K. calculate enthalpy change for the reaction at 298 K.
$N H_{2} C N_{(g)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$