The reaction A+B yields C+D rate=k[A][B]2 has an intial rate of 0.00220Ms what will the initial rate be if [A] is halved and [B] is tripled? Answer must be in Ms What will the intial rate be if [A] is tripled and [B] is halved? Answer must be in Ms

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The reaction  A+B yields C+D  rate=k[A][B]2  has an intial rate of 0.00220Ms  what will the initial rate be if [A] is halved and [B] is tripled? Answer must be in Ms  What will the intial rate be if [A] is tripled and [B] is halved? Answer must be in Ms

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Step 1The reaction is:The rate law for this reaction is written as:r=k[A][B]2Substitute initial rate of the reaction (r)0.0220ms−1 in rate law.0.0220Ms=k[A][B]2An expression for initial rate is obtained using the rate law by substituting the value of initial rate.Step 2When [A] is halved and [B] is tripled, the new concentrations are:[A′]=[A]2 and [B′]=[B]×3The new rate law is written as:r=k[A′][B′]2Substitute, [A]2 for [A&#039;] and 3[B] for [B′] in new rate law as follows:1) r=k[A]2(3[B])2=92k[A][B]2Since,k[A][B]2=0.0220MS−1Substitute, 0.0220ms−1 fpr k[A][B]2 in equation (1), thusr=92×0.0220ms−1=0.099ms−1Rate of the reaction become 0.099ms−1When the initial concentrations of the reactants are changed by any factor, then the first step is to write the new concentrations in terms of the initial concentrations. Then, write the new rate law in terms of the initial concentrations. And then calculate the new rate law by substituting the values accordingly.Step 3When [A] is tripled and [B] is halved, so the new concentrations are:[A″]=3×[A] and [B″]=[B]2The new rate law is written as:r=k[A″][B″]2Substitute, 3[A] for [A′] and [B]2 for [B′] in new rate law as follows:2) r=k(3×[A])([B]2)2=34k[A][B]2Since,k[A][B]2=0.0220Ms−1Substitute, 0.0220ms−1 for k[A][B]2 in equation (2), thus

The reaction A+B\ yields\ C+D rate=k[A][B]^{2} has an in

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