The reaction of cyan-amide- N H 2 CN s - with dioxygen was carried out in a bomb calorimeter- and - U was found to be -742-7kJ mol -1 at 298 K- Calculate enthalpy change for the reaction at 298 K- N H 2 CN g - 3 2 O 2 g - N 2 g - CO 2 g - H 2 O l

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Byju's Answer

Standard XII

Chemistry

Heat of Reaction

The reaction ...

Question

The reaction of cyan-amide, $N H_{2} C N (s)$ , with dioxygen was carried out in a bomb calorimeter, and $Δ U$ was found to be $- 742.7 k J {m o l}^{- 1}$ at 298 K. Calculate enthalpy change for the reaction at 298 K.
$N H_{2} C N (g) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + {C O}_{2} (g) + H_{2} O (l)$

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Solution

${N H_{2} C N}_{(g)} + \frac{3}{2} {O_{2}}_{(g)} ⟶ {N_{2}}_{(g)} + {C O_{2}}_{(g)} + {H_{2} O}_{(l)}$
From the above reaction,
$Δ n_{g} = n_{P} - n_{R} = (1 + 1) - (1 + \frac{3}{2}) = - \frac{1}{2}$
Given:-
$Δ U = - 742.7 k J / m o l$
$T = 298 K$
As we know that,
$Δ H = Δ U + Δ n_{g} R T$
$\Rightarrow Δ H = - 742.7 + (- \frac{1}{2} \times 8.314 \times 10^{- 3} \times 298)$
$\Rightarrow Δ H = - 742.7 - 1.238 = - 743.938$
Hence the enthalpy change for the given reaction is $- 743.938 k J / m o l$ .

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Similar questions

Q. The reaction of cyanamide

N H_{2} C N (s)

with dioxygen was carried out in a bomb calorimeter and

△ U

was found to be

- 742.7 k J m o l^{- 1}

at 298 K.

N H_{2} C N (g) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)

Calculate enthalpy change for the reaction at 298 K?

The reaction of cyanamide, $N H_{2} C N_{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $Δ U$ was found to be $- 742.7 k J m o l^{- 1}$ at 298 K. calculate enthalpy change for the reaction at 298 K.
$N H_{2} C N_{(g)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$