JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 16)
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]
$$N{H_2}C{H_{(S)}} + {3 \over 2}{O_{2(g)}} \to {N_{2(g)}} + {O_{2(g)}} + {H_2}{O_{(I)}}$$
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$$-$$1 K$$-$$1]
Answer
741
Explanation
$$N{H_2}CN(s) + {3 \over 2}{O_2}(g)\buildrel {} \over
\longrightarrow {N_2}(g) + C{O_2}(g) + {H_2}O(l)$$
$$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$$
$$\Delta H = \Delta U + \Delta ng\,RT$$
$$ = - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$$
$$ = - 742.24 + 1.24$$
$$ = -741$$ kJ/mol
$$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$$
$$\Delta H = \Delta U + \Delta ng\,RT$$
$$ = - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$$
$$ = - 742.24 + 1.24$$
$$ = -741$$ kJ/mol
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