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The data given below are for the reactio...

The data given below are for the reactionof NO and `Cl_(2)` to form NOCl at 295 K
`[Cl_(2)]" "[NO]" ""initial rate(mol litre"^(-1)"sec"^(-1))`
`0.05" "0.05" "1xx10^(-3)`
`0.15" "0.05" 3xx10^(-3)`
`0.05" "0.15" "9xx10^(-3)`
If `[Cl_(2)]` is halved and (NO) is double the rate of the reaction is ________

A

halved

B

doubled

C

unchanged

D

becomes 1/4th

Text Solution

Verified by Experts

The correct Answer is:
B
`(r_(2))/(r_(1))=(Kxx(2)^(2)xx(1/2)^(1))/(Kxx(1)^(2)xx(1)^(1)),(r_(2))/(r_(1))=4xx1/2=2impliesr_(2)=2xxr_(1)`
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