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For a reaction NO(g) +O(2) (g) rarr 2NO...

For a reaction `NO_(g) +O_(2) (g) rarr 2NO_(2)(g)`
Rate ` =k[NO^(2)] [O_(2)]` if the volume of the reaction vessel is doubled the rate of the reaction :

A

will diminish to 1/4 of initial value

B

will diminish to 1/8 of initial value will grow 4 times

C

will grow 4times

D

will grow 8 times

Text Solution

Verified by Experts

The correct Answer is:
B
(B) On increasing the volume to twice the value the concentration of each species is reduced by a factor of 2 . Therefore
Rate = `k[NO]^(2) [O_(2)]`
`"Rate"_(2)= k(([NO])/(2))^(2)(([O_(2)])/(2))`
`("Rate"_(2))/("Rate"_(1)) = (1)/(8)`
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