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Standard XII

Chemistry

Introduction

The density o...

Question

The density of an equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ at 101.32 $K P_{a}$ is 3.62 g $d m^{3}$ at 288 K and 1.84 g $d m^{3}$ at 348 K.

What is the heat of the reaction for the following reaction?

$N_{2} O_{4} ⇌ 2 N O_{2} (g)$

Δ_{r} H = 37.29

kJ mol

^{- 1}

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Δ_{r} H = 75.68

kJ mol

^{- 1}

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Δ_{r} H = 95.7

kJ mol

^{- 1}

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Δ_{r} H = 151.3

kJ mol

^{- 1}

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Solution

The correct option is B $Δ_{r} H = 75.68$ kJ mol $^{- 1}$ .
At 288 K, $M_{a v g .} = \frac{3.62 \times 0.0821 \times 288}{1}$
$\frac{92}{M_{a v g .}} = 1 + α \Rightarrow K_{P 1} = \frac{4 α^{2}}{1 - α^{2}}$
Similarly at $348 K, M^{'} / a v g . = \frac{1.84 \times 0.0821 \times 348}{1}$
$\frac{92}{M^{'} a v g} = 1 + α^{'} \Rightarrow K_{P_{2}} = \frac{4 α^{' 2}}{1 - α^{' 2}}$
$l o g \frac{K_{P_{2}}}{K_{P_{1}}} = \frac{Δ H^{o}}{2.303 R} [\frac{1}{288} - \frac{1}{348}]$
so,
$Δ_{r} H = 75.68 k J m o l^{1}$

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Similar questions

Q. The enthalpy changes of the following reactions at

27^{\circ} C

are

$N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)$	$Δ_{r} H = - 411 k J / m o l$
$H_{2} (g) + S (s) + 2 O_{2} (g) \to H_{2} S O_{4} (l)$	$Δ_{r} H = - 811 k J / m o l$
$2 N a (s) + S (s) + 2 O_{2} (g) \to N a_{2} S O_{4} (s)$	$Δ_{r} H = - 1382 k J / m o l$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$	$Δ_{r} H = - 92 k J / m o l;$

from these data, the heat change of reaction at constant volume (in kJ/mol) at

27^{\circ} C

for the process

2 N a C l (s) + H_{2} S O_{4} (l) \to N a_{2} S O_{4} (s) + 2 H C l (g)

Q. For the hypothetical reaction,

A_{2} g + B_{2} (g) ⇌ 2 A B (g)

Δ_{r} G^{\circ} a n d Δ_{r} S^{\circ}

are 20 kj/mol and -20

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

respectively at 200 K

Δ_{r} C_{P} i s 20 J K^{- 1} m o l^{- 1}

then value

\frac{Δ_{r} H^{\circ}}{10}

of (KJ/mol) at 400 K is:

Q. From the following data at

25^{\circ} C

Reaction	$Δ_{r} H^{\circ}$ kJ/mol
$\frac{1}{2} H_{2} (g) + \to \frac{1}{2} O_{2} (g)$	$+ 42$
$H_{2} (g) + \frac{1}{2} O_{2} (g) H_{2} O (g)$	$- 242$
$H_{2} (g) \to 2 H (g)$ $O_{2} (g) \to 2 O (g)$	$+ 436$ $+ 495$

Δ_{r} H

and

Δ_{r} S

for the reaction

M g O (s) + C (s) \to M g (s) + C O (g)

are

491.18

m o l^{- 1}

and

197.67

m o l^{- 1}

respectively. Calculate the temperature at which Gibb's energy change will be zero.

Consider the reaction:

$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g), Δ_{r} H = - 111 k J$ . If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be:

(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

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The density of an equilibrium mixture of N2O4 and NO2 at 101.32 KPa is 3.62 g dm3 at 288 K and 1.84 g dm3 at 348 K. What is the heat of the reaction for the following reaction?N2O4⇌2NO2(g)

The correct option is B ΔrH=75.68 kJ mol−1. At 288 K, Mavg.=3.62×0.0821×288192Mavg.=1+α⇒KP1=4α21−α2Similarly at 348K,M′/avg.=1.84×0.0821×348192M′avg=1+α′⇒KP2=4α′21−α′2logKP2KP1=ΔHo2.303R[1288−1348]so,ΔrH=75.68kJmol1

The density of an equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ at 101.32 $K P_{a}$ is 3.62 g $d m^{3}$ at 288 K and 1.84 g $d m^{3}$ at 348 K.

What is the heat of the reaction for the following reaction?

$N_{2} O_{4} ⇌ 2 N O_{2} (g)$