Question

The equilibrium constant of the equilibrium equation H2O + Co - H2 + CO2 is 0.44 at 1259 k .The value of equilibrium constant H2 + CO will be ?

Answer 1

Answer:

The equilibrium constant for the reaction $${H_2}O + CO - - - > {H_2} + C{O_2}$$ is 0.44 at 1260k. what will be the equibilibrium constant value for the reaction $${H_2} + C{O_2} - - > {H_2}O + CO$$

Answer 2

Explanation:

The given reaction is :-

CH

3

COOH(l)+C

2

H

5

OH(l)⇌H

3

COOC

2

H

5

(l)+H

2

O(l)

Initial conc. : 4 4 0 0

At eqm : 4−x 4−x x x

Given, initial conc. of acid & alcohol is & moles / litre. Let, x moles of each acid & alcohol has reacted.

$$\Rightarrow \quad { K }_{ C }=\dfrac { \left[ { H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } \right] \left[ { H }_{ 2 }O \right] }{ \left[ { CH }_{ 3 }COOH \right] \left[ { C }_{ 2 }{ H }_{ 5 }OH \right] } $$

$$\Rightarrow \quad 4=\dfrac { x.x }{ \left( 4-x \right) \left( 4-x \right) } $$

$$\Rightarrow \quad \dfrac { { x }^{ 2 } }{ { \left( 4-x \right) }^{ 2 } } =4\quad \quad \quad \Rightarrow \dfrac { x }{ 4-x } =2$$ (Taking square root)

⇒

4−x

x

=2⇒x=8−2x

⇒ x=8/3

So, no. of moles of acid at equilibrium =4−

3

8

=

3

4

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