The rate law for the reaction RCl+NaOH(aq)→ROH+NaCl is given by Rate = k[RCl]. The rate of the reaction will be

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unaffected by increasing temperature of the reaction

doubled on doubling the concentration of NaOH

halved on reducing the concentration of NaOH to one half

halved on reducing the concentration of RCI to half

answer is D.

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Detailed Solution

R1=k[RC]1[NaOH]0R2=k[0.5 RCl]1∴ R2=0.5 R1

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