The reaction of cyanamide, NH 2 CN ( s ) , with oxygen was run in a bomb calorimeter and Δ E was found to be − 742.7 kJ / mol of NH 2 CN ( s ) at 298 K . Calculate Δ H 298 for the reaction, N H

Question

The reaction of cyanamide, NH 2 ​ CN ( s ) , with oxygen was run in a bomb calorimeter and Δ E was found to be − 742.7 kJ / mol of NH 2 ​ CN ( s ) at 298 K . Calculate Δ H 298 ​ for the reaction, N H 2 ​ CN ( s ) + 2 3 ​ O 2 ​ ( g ) ⟶ N 2 ​ ( g ) + C O 2 ​ ( g ) + H 2 ​ O ( l )

TestHub · Accepted Answer

The number of moles of gaseous reactant (O₂) is $\frac{3}{2}$ , and the number of moles of gaseous products (N₂ + CO₂) is 2. Therefore, $Δ H = Δ E + Δ (P V) = Δ E + (Δ n) RT$ .

$Δ n = 2 - \frac{3}{2} = \frac{1}{2}$

$Δ H = - 742.7 k J + (0.500 m o l) (8.314 \times 1 0^{- 3} k J m o l^{- 1} K^{- 1}) (298 K)$

$Δ H = - 742.7 k J + 1.24 k J = - 741.46 k J \approx - 741.5 k J$

Chemistry - Thermochemistry Question with Solution | TestHub

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