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Standard XII

Chemistry

Enthalpy of Solution

The reaction ...

Question

The reaction $N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ was carried out at 300K in a bomb
calorimeter. The heat released was $743 k j m o l^{- 1}$ . The value of $△ H_{300 K}$ for this reaction be:-

- 740.5 k j m o l^{- 1}

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- 741.75 k j m o l^{- 1}

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- 743.0 k j m o l^{- 1}

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- 744.25 k j m o l^{- 1}

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Solution

The correct option is B $- 740.5 k j m o l^{- 1}$
In a bomb calorimeter, heat released $= - Δ U$
$∴ U = - 743 K J m o l^{- 1}$
$∴ Δ H = Δ U - Δ n g R T$
where, $Δ n g =$ difference between gaseous moles of products and reactants
$∴ Δ n g = (1 + 1) - \frac{3}{2} = \frac{1}{2}$
$∴ Δ H = Δ U + \frac{1}{2} R T$
$= - 743 + \frac{1}{2} \times 8.314 \times 300 \times 10^{- 3}$
$Δ H_{100} = - 741.75 K H m o l^{- 1}$

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Similar questions

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The reaction NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l) was carried out at 300K in a bombcalorimeter. The heat released was 743 kj mol−1. The value of △H300K for this reaction be:-

The correct option is B −740.5 kj mol−1In a bomb calorimeter, heat released =−ΔU∴U=−743 KJ mol−1∴ΔH=ΔU−ΔngRTwhere, Δng= difference between gaseous moles of products and reactants∴Δng=(1+1)−32=12∴ΔH=ΔU+12RT=−743+12×8.314×300×10−3ΔH100=−741.75 KH mol−1

The reaction $N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ was carried out at 300K in a bomb
calorimeter. The heat released was $743 k j m o l^{- 1}$ . The value of $△ H_{300 K}$ for this reaction be:-