The reaction of cyanamide, NH _ 2 CN _ (s) with oxygen was run in a bomb calorimeter and Δ U was found to be - 742 . 24kJ mol ^ - 1 . The magnitude of Δ H _ 298 for the reaction NH _ 2 CN _ (s) + (3)/(2) O _ 2(g) → N _

Correct answer: 0 Δ H = Δ U + Δ n _ g RT = - 742 . 24 + (1)/(2) × (8 . 314)/(1000) × 298 = - 741kJ / mol Hence answer is (741)

JEE Main 202125 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2021 · 25 February · Morning Shift · Chemistry · Thermodynamics (C)

The reaction of cyanamide, NH _ 2 CN _ (s) with oxygen was run in a...

The reaction of cyanamide,

{NH}_{2}{CN}_{\left(s\right)}

with oxygen was run in a bomb calorimeter and

\Delta U

was found to be

- 742 . 24kJ{mol}^{- 1}

. The magnitude of

{\Delta H}_{298}

for the reaction

{NH}_{2}{CN}_{\left(s\right)} + \dfrac{3}{2}{O}_{2\left(g\right)} \to {N}_{2\left(g\right)} + {O}_{2\left(g\right)} + {H}_{2}{O}_{\left(l\right)}

kJ

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

R = 8 . 314J{mol}^{- 1}{K}^{- 1}

]

Correct answer

Step-by-step solution

\Delta H = \Delta U + {\Delta n}_{g}RT $$ = - 742 . 24 + \dfrac{1}{2} \times \dfrac{8 . 314}{1000} \times 298 $$ = - 741kJ / mol $Hence answer is \left(741\right)$

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