The equilibrium constant KP for the reaction H2g - CO2g - H2Og - COg is 4-0 at 1660-C- Initially 0-80 mole H2 and 0-80 mole CO2 are injected into a 5-0 litre flask- What is the equilibrium concentration of CO2g-

The correct option is B 0.0534 M 0.8 x H_2(g) 0.8 x #160; CO_2(g) ⇌ x H_2O(g) + x CO(g) moles of eqsn #160;conc. at eqm 0.8 x5(0.8 x5) ...

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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

The equilibri...

Question

The equilibrium constant $K_{P}$ for the reaction $H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)$ is $4.0$ at $1660^{\circ} C$ . Initially $0.80 m o l e H_{2}$ and $0.80 m o l e C O_{2}$ are injected into a $5.0 l i t r e$ flask. What is the equilibrium concentration of $C O_{2} (g)$ ?

0.533 M

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0.0534 M

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5.34 M

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Solution

The correct option is B $0.0534 M$
$H_{2} (g) 0.8 - x C O_{2} (g) 0.8 - x ⇌ H_{2} O (g) x + C O (g) x$
moles of eqsn
conc. at eqm $\frac{0.8 - x}{5} (\frac{0.8 - x}{5}) (\frac{x}{5}) (\frac{x}{5})$
$∵ Δ n_{g} = 0$
$∴ K_{c} = \frac{{(\frac{x}{5})}^{2}}{{(\frac{0.8 - x}{5})}^{2}}$
$\Rightarrow 2 = \frac{x}{0.8 - x}$
$\Rightarrow x = 0.533$
$[C O_{2} (g)] = \frac{0.8 - 0.533}{5}$

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Similar questions

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

Q. The equilibrium constant for the reaction

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

is 3 at 500K. In a 2 litre vessel 60gm of water gas [equimolar mixture of CO(g) and

H_{2} (g)

] and 90 gm of steam is initially taken.

What is the equilibrium concentration of

H_{2} (g)

at equilibrium (mole/L)?

Q. The equilibrium constant of the equation,

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

0.44

1259 K

. The value of equilibrium constant of the equation,

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

will be:

Q. At a given temperature,

K c

4

for the reaction :

H_{2 (g)} + C {O_{2}}_{(g)} \Leftrightarrow H_{2} O_{(g)} + C O_{(g)}

Initially

0.6

moles each of

H_{2}

and

C O_{2}

are taken in

1

liter flask. The equilibrium concentration of

H_{2} O_{(g)}

is :

Q. For the reaction,

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

, if the initial concentration of

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and

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H_{2}

is consumed at equilibrium, the correct expression of

K_{p}

is:

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The equilibrium constant KP for the reaction H2(g)+CO2(g)⇌H2O(g)+CO(g) is 4.0 at 1660∘C. Initially 0.80 mole H2 and 0.80 mole CO2 are injected into a 5.0 litre flask. What is the equilibrium concentration of CO2(g)?

The correct option is B 0.0534 MH2(g)0.8−xCO2(g)0.8−x⇌H2O(g)x+CO(g)xmoles of eqsn conc. at eqm 0.8−x5(0.8−x5)(x5)(x5)∵Δng=0∴Kc=(x5)2(0.8−x5)2⇒2=x0.8−x⇒x=0.533[CO2(g)]=0.8−0.5335

The equilibrium constant $K_{P}$ for the reaction $H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)$ is $4.0$ at $1660^{\circ} C$ . Initially $0.80 m o l e H_{2}$ and $0.80 m o l e C O_{2}$ are injected into a $5.0 l i t r e$ flask. What is the equilibrium concentration of $C O_{2} (g)$ ?