The reaction of cyanamide NH2CNs with dioxygen was carried out in a bomb calorimeter and - U was found to be -742-7 kJmol-1 at 298 K- Calculate enthalpy change for the following reaction at 298 K- NH2CNs-3-2O2g- N2g-CO2g-H2Ol

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Standard XII

Chemistry

Enthalpy of Solution

The reaction ...

Question

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 following reaction at $298 K$ .

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

+ 391.7

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- 24.8

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- 602.0

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- 741.5

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Solution

The correct option is C $- 741.5$ kJ
As we know,
$Δ n = 1 + 1 - 1.5 = 0.5$

Now,
$Δ H = Δ U + Δ n R T$

$Δ H = - 742.7 + 0.5 \times 8.314 \times 10^{- 3} \times 292$

$Δ H = - 742.7 + 1.2$

$Δ H = - 741.5 k J$

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Similar questions

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.

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

Q. The reaction of cyanamide

(N H_{2} C N)

with dioxygen was carried out in a bomb calorimeter and

Δ U

was found to be

- 742.7 {kJmol}^{- 1}

298 K

. Calculate the enthalpy change for the reaction at

298 K

N H_{2} C N (s) + \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?

Q. The reaction of cyan-amide,

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. When the following reaction was carried out in a bomb calorimeter,

Δ E

is found to be

742.7 k J / m o l

N H_{2} C N (s)

298 K .

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

Calculate

Δ H_{298}

for the reaction.

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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 kJmol−1 at 298K. Calculate enthalpy change for the following reaction at 298K. NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l)

The correct option is C −741.5 kJAs we know,Δn=1+1−1.5=0.5Now,ΔH=ΔU+ΔnRTΔH=−742.7+0.5×8.314×10−3×292ΔH=−742.7+1.2ΔH=−741.5 kJ

The correct option is C $- 741.5$ kJ
As we know,
$Δ n = 1 + 1 - 1.5 = 0.5$

Now,
$Δ H = Δ U + Δ n R T$

$Δ H = - 742.7 + 0.5 \times 8.314 \times 10^{- 3} \times 292$

$Δ H = - 742.7 + 1.2$

$Δ H = - 741.5 k J$