The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and $\Delta$U was found to be $-$742.24 kJ mol$-$1. The magnitude of $\Delta$H298 for the reaction
$$N{H_2}C{H_{(S)}} + {3 \over 2}{O_{2(g)}} \to {N_{2(g)}} + {O_{2(g)}} + {H_2}{O_{(I)}}$$
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$-$1 K$-$1]
Answer (integer)
741
Solution
$$N{H_2}CN(s) + {3 \over 2}{O_2}(g)\buildrel {} \over
\longrightarrow {N_2}(g) + C{O_2}(g) + {H_2}O(l)$$<br><br>$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$<br><br>$\Delta H = \Delta U + \Delta ng\,RT$<br><br>$= - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$<br><br>$= - 742.24 + 1.24$<br><br>$= -741$ kJ/mol
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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