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Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is D $- 741.75 k J m o l^{- 1}$
$N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$

Heat released $= - 743 k J / m o l$

$△ H_{300 K} = ? △ H = △ U + △ n_{g} R T$

$△ U =$ change in internal energy

$△ n_{g} =$ change in number of moles

For above reaction,

$△ n_{g} = Σ n_{g} (p r o d u c t) - Σ n_{g} (r e a c t a n t)$

$= 2 - 1.5 = 0.5$

$△ U = - 743 k J / m o l$

$T = 300 K$

$R = 8.314 \times 10^{- 3} k J {m o l}^{- 1} K^{- 1}$

$△ H = (- 743) + (0.5) (8.314 \times 10^{- 3}) (300)$

$= - 743 + 1.2471 = - 741.7529 k J {m o l}^{- 1}$

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Similar questions

Q. 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}

△ H_{300 K}

for this reaction would be_________.

Q. 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
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743 k j m o l^{- 1}

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

This reaction is carried out in a bomb calorie-meter. The heat released was

743 K J {m o l}^{- 1}

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The reaction of cyanamide, NH₂CN_(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol^–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

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N H_{2} C N (s)

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