JEE PYQ: Thermodynamics Question 3
Question 3 - 2021 (25 Feb Shift 1)
The reaction of cyanamide, $\text{NH}2\text{CN}{(s)}$, with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24$ kJ mol$^{-1}$. The magnitude of $\Delta H_{298}$ for the reaction $\text{NH}2\text{CN}{(s)} + \frac{3}{2}\text{O}_2(g) \to \text{N}_2(g) + \text{CO}_2(g) + \text{H}_2\text{O}(\ell)$ is ___ kJ. (Rounded off to the nearest integer)
[Assume ideal gases and $R = 8.314$ J mol$^{-1}$ K$^{-1}$]
Show Answer
Answer: 741
Solution
$\Delta n_g = (1+1) - \frac{3}{2} = \frac{1}{2}$. $\Delta H = \Delta U + \Delta n_g RT = -742.24 + \frac{1}{2} \times \frac{8.314 \times 298}{1000} = -742.24 + 1.24 = -741$ kJ/mol. Magnitude = 741.
BETA
