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Rate law for a gaseous reaction A + B ...

Rate law for a gaseous reaction
`A + B rarr C + D` is given by
Rate `= K[A]^(2)[B]^(0)`
The volume of reaction vessel containing these gases is suddenly reduced to one fourth the volume. The rate of reaction relative to original rate would be

A

`(1)/(16)`

B

`(16)/(1)`

C

`(1)/(8)`

D

`(8)/(1)`

Text Solution

Verified by Experts

The correct Answer is:
B
On reducing the volume to `1//4th` increaeses the concentration of each reacting species increases four times. As rate is proportional to square of concentration of `A` and is independent of concentration of `B` rate increases sixteen times.
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