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Standard XII

Chemistry

Effective Equilibrium Constant

Carbon monoxi...

Question

Carbon monoxide in water gas, reacts with steam according to the reaction:
$C O (g) + H_{2} O (g) \to C O_{2} (g) + H_{2} (g); K e q = 1.0$
If three volumes of water gas (containing 1:2 of $C O$ and $H_{2}$ ) are mixed with 3 volumes of steam, then the volume ratio of the four gases present at equilibrium is

1 : 5 : 1 : 5

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1 : 4 : 3 : 1

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1 : 5 : 1 : 1

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1 : 1 : 1 : 2

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Solution

The correct option is B $1 : 5 : 1 : 5$
$C O (g) + H_{2} O (g) \to C O_{2} (g) + H_{2} (g)$ ;

$\begin{matrix} 1 & 3 & 0 & 2 1 - x & 3 - x & x & 2 + x \end{matrix}$

$K_{e q} = \frac{x (2 + x)}{(1 - x) (3 - x)} = 1$

$x^{2} + 2 x = 3 - 4 x + x^{2}$

$x = 0.5$
Hence, the ratio for the gases, $C O (g) : H_{2} O (g) : C O_{2} (g) : H_{2} (g) = 0.5 : 2.5 : 0.5 : 2.5$
Or $1 : 5 : 1 : 5$
Option A is correct.

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Similar questions

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

Δ H = 241.8 k J

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

Δ H = 283 k J

The heat evolved during the combustion of

112

litre of water gas (mixture of equal volume of

H_{2}

and

C O

) is:

Q. A sample of water gas has a composition by volume of 50%

H_{2}

, 45%

C O

and 5%

C O_{2}

. Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of

H_{2}

. The equation for the reaction is:

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

Q. The heat evolved during the combustion of 112 litre of water gas at STP (mixture of equal volume of

H_{2}

and CO) is :

Given:

H_{2} (g) + 1 / 2 O_{2} (g) = H_{2} O (g); Δ H = - 241.8 k J

C O (g) + 1 / 2 O_{2} (g) = C O_{2} (g); Δ H = - 283 k J

1

mole of

H_{2} O

and

1

mole of

C O

are heated in a

10

litre vessel at

1259 K

40 % H_{2} O

reacts with

C O

according to the following equations,

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

;
Calculate equilibrium constant for the reaction.

Three containers have their volumes in ratio 3:4:5. They are full of mixtures of milk and water in the ratio of (4:1),(3:1) and (5:2) respectively . The ratio of milk and water in the fourth container contains all the above three mixtures is=

Options....

(1.)4:1. (2.)151:48

(3.)157:53. (4.)5:2

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Carbon monoxide in water gas, reacts with steam according to the reaction:CO(g)+H2O(g)→CO2(g)+H2(g);Keq=1.0If three volumes of water gas (containing 1:2 of CO and H2) are mixed with 3 volumes of steam, then the volume ratio of the four gases present at equilibrium is

The correct option is B 1:5:1:5CO(g)+H2O(g)→CO2(g)+H2(g);13021−x3−xx2+xKeq=x(2+x)(1−x)(3−x)=1x2+2x=3−4x+x2x=0.5Hence, the ratio for the gases, CO(g):H2O(g):CO2(g):H2(g)=0.5:2.5:0.5:2.5Or 1:5:1:5Option A is correct.