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The equlibrium constant is 6.0xx10^(-4) ...

The equlibrium constant is `6.0xx10^(-4)` for the `N_(2)+O_(2)hArr2NO` reaction. If the concentration of nitrogen is `0.10` mol/L and concentration of oxygen is 0.20 mol/L at equilibrium. Then the concentration of nitric oxide at equlibrium is

A

`10.9xx10^(-3)mol//L`

B

`1.09xx10^(-3)mol//L`

C

`10.9xx10^(-5)mol//L`

D

`1.09xx10^(-5)mol//L`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(c)=([NO]^(2))/([N_(2)][O_(2)])`
`6xx10^(-4)=([NO]^(2))/(0.10xx0.20`
`[NO]=sqrt(6.0xx10^(-4)xx0.10xx0.20)`
`=1.09xx10^(-3)mol//L`
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