Find the rate law for the reaction, {\CHCl}_{3( {\g})}+ {\Cl}_{2( {\g})} \rightarrow {\CCl}_{4( {\g})}+ {\HCl}_{( {\g})} if order of reaction with respect to {\CHCl}_{ {\a}( {\g})} is one and \frac{1}{2} with {\Cl}_{2( {\g})}.

Correct Answer is : Rate =k[CHCl3][Cl2]1∕2

Rate = {\k} [ {\CHCl}_3] [ {\Cl}_2] ^{1 \/ 2}

Rate = {\k} [ {\CHCl}_3] ^2 [ {\Cl}_2] ^{1 \/ 2}

Rate = {\k} [ {\CHCl}_3] ^{3 \/ 2} [ {\Cl}_2]

Rate = {\k} [ {\CHCl}_3] ^{1 \/ 2} [ {\Cl}_2]

MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Chemistry

Question : 35 of 150

Marks: +1, -0

Find the rate law for the reaction, CHCl3(g)+Cl2(g)⟶CCl4(g)+HCl(g) if order of reaction with respect to CHCla(g) is one and ‌

with Cl2(g).

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