The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{s})}$ with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The magnitude of $\Delta \mathrm{H}_{298}$ for the reaction $$\mathrm{NH}_{2} \mathrm{CN}_{(\mathrm{S})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}$$ is _____ kJ. (Rounded off to the nearest integer) [Assume ideal gases and $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$]

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Chemical Thermodynamics

The reaction of cyanamide,

NH_{2} CN_{(s)}

with oxygen was run in a bomb calorimeter and

Δ U

was found to be

- 742.24 kJ mol^{- 1}

. The magnitude of

Δ H_{298}

for the reaction

NH_{2} CN_{(S)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + CO_{2 (g)} + H_{2} O_{(l)}

is _____ kJ. (Rounded off to the nearest integer) [Assume ideal gases and

R = 8.314 J mol^{- 1} K^{- 1}

]

Answer: 741

A bomb calorimeter works at constant volume, so the heat it measures is

Δ U

, and

Δ H = Δ U + Δ n_{g} R T

Only gaseous species enter

Δ n_{g}

: two moles of gas are formed, nitrogen and carbon dioxide, against

\frac{3}{2}

mole of oxygen consumed, the solid and the liquid counting for nothing, so

Δ n_{g} = 2 - \frac{3}{2} = + 0.5

. Hence

Δ H = - 742.24 + 0.5 \times 8.314 \times 1 0^{- 3} \times 298 = - 742.24 + 1.24 = - 741.00 kJ mo l^{- 1}

The magnitude of

Δ H_{298}

is 741 kJ.

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Integer answer