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Calorimetry

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 $300 k$ in a bomb calorimeter. The heat released was $742 k J m o l^{- 1}$ . The value of $△ H_{300 K}$ for this reaction would 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 C $- 741.75 k J m o l^{- 1}$
Enthalpy change for a reaction $(Δ H)$ is given by the expression,
$Δ H = Δ U + Δ n_{g} R T$
where,
$Δ U =$ change in internal energy
$Δ n_{g} =$ change in number of moles

For the given reaction,

$Δ n_{g} = \sum n_{g} (p r o d u c t - \sum n_{g} (r e a c t a n t)$

$= (2 - 1.5) m o l e s$

$Δ n_{g} = 0.5 m o l e s$ ,

$Δ U = - 742.7 k J / m o l$

$T = 298 k$

$R = 8.314 \times 10^{- 3} k J / m o l$

Substituting the values in the expression of $Δ H$

$Δ H = (- 742.7) + (0.5) (298) (8.314 \times 10^{- 3})$
$= - 742.7 + 1.2$
$= - 741.5 k J / m o l$

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