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At 700 K, CO(2) and H(2) react to form C...

At `700 K, CO_(2)` and `H_(2)` react to form `CO` and `H_(2)O`. For this purpose, `K_(c )` is `0.11`. If a mixture of `0.45` mol of `CO_(2)` and `0.45` mol of `H_(2)` is heated to `700 K`.
(a) Find out amount of each gas at equilibrium.
(b) When equilibrium has been reached, another `0.34` mol of `CO_(2)` and `0.34` mol of `H_(2)` are added to the reaction mixture. Find the composition of of mixture at new equilibrium.

A

`[CO]=[H_(2)O]=0.224` moles

B

`[CO]=0.112` moles, `[H_(2)O]=0.224` moles

C

`[CO]=0.224` moles, `[H_(2)O]=0.112` moles

D

`[CO]=[H_(2)O]=0.112` moles

Text Solution

Verified by Experts

The correct Answer is:
D
The given equilibrium is
`{:(,CO_(2)(g),+,H_(2)(g) , hArr, CO(g),+,H_(2)O(g)),("At "t=0,0.45,,0.45,,0,,0),("At equilibrium",(0.45-x),,(0.45-x),,x,,x):}`
`K_(c)=(x^(2))/((0.45-x)^(2)) rArr 0.11 = (x^(2))/((0.45-x)^(2)) rArr x=0.112`
`therefore [CO_(2)]=[H_(2)]=0.45-x=(0.45-0.112)=0.3379=0.34` mole each
`[CO]=[H_(2)O]=x=0.112` mole each
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