For a given reaction-C s - O 2 g - Co g - H 2 g at 25- C - H -131 kJ - molGiven- K p -9-36 - 10-17Calculate the value of reaction constant at 700- C-A- 0-915B- 0-214C- 0-583D- 0-797

The correct option is D 0.797As per van't hoff equation, log K_2K_1 = Δ H2.303R× nbsp; 1T_1 1T_2 Given: K_1 =9.36 × 10^ 17 nbsp; Δ H = 131 kJ/mole T_1 = ...

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Standard XII

Chemistry

Le Chatelier's Principle for Del N Greater Than Zero

For a given r...

Question

For a given reaction:
$C (s) + O_{2} (g) \to C o (g) + H_{2} (g)$ at $25^{o} C$ $Δ H = 131 k J / m o l$
Given: $K_{p}$ = $9.36 \times 10^{- 17}$
Calculate the value of reaction constant at $700^{o} C$ .

0.915

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0.214

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0.583

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0.797

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Solution

The correct option is D $0.797$
As per van't hoff equation,
$l o g \frac{K_{2}}{K_{1}} = \frac{Δ H}{2.303 R} \times \frac{1}{T_{1}} - \frac{1}{T_{2}}$
Given: $K_{1}$ = $9.36 \times 10^{- 17}$
$Δ H = 131 k J / m o l e$ $T_{1} = 298 K$ $T_{2} = 973 K$
put up the values in equation, we get
$l o g \frac{K_{2}}{9.36 \times 10^{- 17}} = \frac{131}{2.303 R} \times \frac{1}{298} - \frac{1}{973}$
On solving,
$K_{2} = 0.797$

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Similar questions

Q. For a given reaction:

C (s) + O_{2} (g) \to C o (g) + H_{2} (g)

25^{o} C

Δ H = 131 k J / m o l

Given:

K_{p}

9.36 \times 10^{- 17}

Calculate the value of reaction constant at

700^{o} C

Q. Calculate

Δ H

k J

for the following reaction:

C (g) + O_{2} (g) \to C O_{2} (g)

Given that,

H_{2} O (g) + C (g) \to C O (g) + H_{2} (g)

;

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

;

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

;

Δ H = - 242 k J

Q. Equilibrium constants for the following reaction at 1200 K are given.

2 H_{2} O (g) ⇋ 2 H_{2} (g) + O_{2} (g); K_{1} = 6.4 \times 10^{- 8}

2 C O_{2} (g) ⇋ 2 C O (g) + O_{2} (g); K_{2} = 1.6 \times 10^{- 6}

The equilibrium constant for the reaction

H_{2} (g) + C O_{2} (g) ⇋ C O (g) + H_{2} O (g)

at 1200 K will be:

Q. The equilibrium constants for the following reactions at

1400 K

are given:

2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g); K_{1} = 2.1 \times 10^{- 13}

2 {C O}_{2} (g) ⇌ 2 C O (g) + O_{2} (g); K_{2} = 1.4 \times 10^{- 12}

Then, the equilibrium constant

K

for the reaction,

H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g)

, is:

Q. Calculate the equilibrium constant for the reaction

H_{2 (g)} + C O_{2 (g)} ⇌ H_{2} O_{(g)} + C O_{(g)}

1395 K

, if the equilibrium constants at

1395 K

for following are:

2 H_{2} O_{(g)} ⇌ 2 H_{2} + O_{2 (g)}; K_{1} = 2.1 \times 10^{- 13}

2 C O_{2 (g)} ⇌ 2 C O_{(g)} + O_{2 (g)}; K_{2} = 1.4 \times 10^{- 12}

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Le Chatelier's Principle for Del N Greater Than Zero

Standard XII Chemistry

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For a given reaction: C(s)+O2(g)→Co(g)+H2(g) at 25oC ΔH=131kJ/mol Given: Kp = 9.36×10−17 Calculate the value of reaction constant at 700oC.

The correct option is D 0.797As per van't hoff equation, logK2K1=ΔH2.303R×1T1−1T2 Given: K1 =9.36×10−17 ΔH=131 kJ/mole T1=298K T2=973K put up the values in equation, we get logK29.36×10−17=1312.303R×1298−1973 On solving, K2=0.797

For a given reaction:
$C (s) + O_{2} (g) \to C o (g) + H_{2} (g)$ at $25^{o} C$ $Δ H = 131 k J / m o l$
Given: $K_{p}$ = $9.36 \times 10^{- 17}$
Calculate the value of reaction constant at $700^{o} C$ .