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A mixture of 1.57 mol of N(2), 1.92 mol ...

A mixture of `1.57 mol` of `N_(2), 1.92 mol` of `H_(2)` and `8.13 mol` of `NH_(3)` is introduced into a `20 L` reaction vessel at `500 K`. At this temperature, the equilibrium constant `K_(c )` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.7xx10^(2)`. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

A

Forward

B

Backward

C

At equilibrium

D

Data is insufficient

Text Solution

Verified by Experts

The correct Answer is:
B
The reaction is : `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`
`Q_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`=(((8.13)/(20)"mol L"^(-1))^(2))/(((1.57)/(20)"mol L"^(-1))((1.92)/(20)"mol L"^(-1))^(3))=2.38xx10^(3)`
As `Q_(c)!=K_(c)`, the reaction mixture is not in equilibrium.
As `Q_(c)gtK_(c)`, the net reaction will be in the backward direction.
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