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Standard XII

Physics

Calorimetry

NH 2 CN s + ...

Question

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

- 740 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 $- 741.75 K J {m o l}^{- 1}$
In a bomb calorimeter, heat released= $- Δ U$
$∴ Δ U = - 743 K J m o l^{- 1}$
$∴ Δ H = Δ U - Δ n g R T$
where, $Δ n g$ = difference between gaseous moles of products and reactants
$∴ Δ n g = (1 + 1) - \frac{3}{2} = \frac{1}{2}$
$∴ Δ H = Δ U + \frac{1}{2} R T$
$= - 743 + \frac{1}{2} \times 8.314 \times 300 \times 10^{- 3}$
$Δ H_{300} = - 741.75 K J m o l^{- 1}$

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Similar questions

Q. The reaction

{N H}_{2} C N (s) + \frac{3}{2} O_{2} (g) + {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}

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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)

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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

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The reaction of cyanamide, $N H_{2} C N_{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $Δ U$ was found to be $- 742.7 k J m o l^{- 1}$ at 298 K. calculate enthalpy change for the reaction at 298 K.
$N H_{2} C N_{(g)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$

Q. The reaction of cyanamide

N H_{2} C N (s)

with dioxygen was carried out in a bomb calorimeter and

△ U

was found to be

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

at 298 K.

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

Calculate enthalpy change for the reaction at 298 K?

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NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l)This reaction is carried out in a bomb calorie-meter. The heat released was 743 KJ mol−1. The value of ΔH300 for this reaction would be:

The correct option is B −741.75 KJ mol−1In a bomb calorimeter, heat released= −ΔU∴ΔU=−743KJmol−1∴ΔH=ΔU−ΔngRTwhere, Δng= difference between gaseous moles of products and reactants∴Δng=(1+1)−32=12∴ΔH=ΔU+12RT=−743+12×8.314×300×10−3ΔH300=−741.75KJmol−1

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