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Standard XII

Chemistry

Enthalpy of Solution

When the foll...

Question

When the following reaction was carried out in a bomb calorimeter, $Δ U$ is found to be $1742.7 k J / m o l$ of $N H_{2} C N (s)$ at 300 K.
${N H}_{2} C N (s) + \frac{3}{2} O_{2} (g) ⟶ N_{2} (g) + {C O}_{2} (g) + H_{2} O (l)$
Calculate ${Δ H}_{3000 K}$ for the reaction . $(R = 8 J / m o l e - K)$ .

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Solution

Given, $Δ U = 1742.7 K J / m o l = 1742.7 \times 10^{3} J / m o l$
$R = 8 J / m o l K, T = 300 K, △ n_{g} = 1 + 1 - \frac{3}{2} = \frac{1}{2}$
$Δ H_{300 K} = (1742.7 \times 10^{3} J / m o l e) + \frac{1}{2} \times 8 J / m o l K \times 300 K$
$Δ H_{300 K} = 1743.9 K J / m o l e$

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When the following reaction was carried out in a bomb calorimeter,ΔU is found to be 1742.7kJ/mol of NH2CN(s) at 300 K.NH2CN(s)+32O2(g)⟶N2(g)+CO2(g)+H2O(l) Calculate ΔH3000K for the reaction .(R=8J/mole−K).

Given, ΔU=1742.7 KJ/mol=1742.7×103 J/molR=8 J/mol K, T=300 K, △ng=1+1−32=12ΔH300 K=(1742.7×103 J/mole)+12×8 J/mol K×300 KΔH300K=1743.9 KJ/mole

Given, $Δ U = 1742.7 K J / m o l = 1742.7 \times 10^{3} J / m o l$
$R = 8 J / m o l K, T = 300 K, △ n_{g} = 1 + 1 - \frac{3}{2} = \frac{1}{2}$
$Δ H_{300 K} = (1742.7 \times 10^{3} J / m o l e) + \frac{1}{2} \times 8 J / m o l K \times 300 K$
$Δ H_{300 K} = 1743.9 K J / m o l e$