1.

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 ________

Answer»

halved
doubled
unchanged
becomes 1/4th

Solution :`(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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