When the following reaction was carried out in a bomb calorimeter- - E is found to be 742-7 kJ-mol of NH2CN s at 298 K- NH2CNs-3-2O2g- N2g-CO2g-H2OlCalculate - H298 for the reaction-

The correct option is A 741.5 kJ NH_4CN(s)+3/2O_2(g)→ N_2(g)+CO_2(g)+H_2O(l) Δ H_298=Δ E+Δ n_gRT= 742.7+1/2× 8.314× 298 = 742.7+1.239 ...

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Byju's Answer

Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

When the foll...

Question

When the following reaction was carried out in a bomb calorimeter, $Δ E$ is found to be $742.7 k J / m o l$ of $N H_{2} C N (s)$ at $298 K .$
$N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$
Calculate $Δ H_{298}$ for the reaction.

- 741.5 k J

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+ 741.5 k J

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- 743.9 k J

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None of these

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Solution

The correct option is A $- 741.5 k J$
$N H_{4} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$
$Δ H_{298} = Δ E + Δ n_{g} R T = - 742.7 + \frac{1}{2} \times 8.314 \times 298$
$= - 742.7 + 1.239$
$Δ H_{298} = - 741.46$

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When the following reaction was carried out in a bomb calorimeter, ΔE is found to be 742.7kJ/mol of NH2CN(s) at 298K.NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l)Calculate ΔH298 for the reaction.

The correct option is A −741.5kJNH4CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l)ΔH298=ΔE+ΔngRT=−742.7+12×8.314×298=−742.7+1.239ΔH298=−741.46

When the following reaction was carried out in a bomb calorimeter, $Δ E$ is found to be $742.7 k J / m o l$ of $N H_{2} C N (s)$ at $298 K .$
$N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$
Calculate $Δ H_{298}$ for the reaction.

The correct option is A $- 741.5 k J$
$N H_{4} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$
$Δ H_{298} = Δ E + Δ n_{g} R T = - 742.7 + \frac{1}{2} \times 8.314 \times 298$
$= - 742.7 + 1.239$
$Δ H_{298} = - 741.46$