A) rate\[\text{mo}{{\text{l}}^{\text{-1}}}\]
B) rate\[\text{mo}{{\text{l}}^{\text{-1}}}\]
C) rate\[\text{mo}{{\text{l}}^{\text{-1}}}\]
D) rate\[\text{mo}{{\text{l}}^{\text{-1}}}\]
Correct Answer: B
Solution :
For the reaction, \[\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,=\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,+\text{H}-\xrightarrow[{}]{{}}\text{H}-\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-\text{H}\]Products On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore Rat\[\text{mo}{{\text{l}}^{\text{-1}}}\] ...(i) Let initially rate law is Rate\[\text{mo}{{\text{l}}^{\text{-1}}}\] ...(ii) If concentration of A and B both are doubled, the rate gets changed by a factor of 8. \[\text{mo}{{\text{l}}^{\text{-1}}}\] ?(iii) \[\text{mo}{{\text{l}}^{\text{-1}}}\] Dividing Eq. (iii) by Eq. (ii), \[1.77\times {{10}^{-5}}\] \[5.65\times {{10}^{-10}}\] \[6.50\times {{10}^{-12}}\]\[5.65\times {{10}^{-13}}\] Hence, rate law is, rate\[5.65\times {{10}^{-12}}\]
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