The instantaneous rate of disappearance of Mn | vedclass

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Question

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$.Given that the

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$1.14 	imes 10^{-2} \ Ms^{-1}$

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(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$.Given that the rate of disappearance of $MnO_4^-$ is $-\frac{d[MnO_4^-]}{dt} = 4.56 	imes 10^{-3} \ Ms^{-1}$.Substituting this into the rate expression: $\frac{1}{2} 	imes (4.56 	imes 10^{-3}) = \frac{1}{5} \frac{d[I_2]}{dt}$.Therefore,the rate of appearance of $I_2$ is $\frac{d[I_2]}{dt} = \frac{5}{2} 	imes 4.56 	imes 10^{-3} \ Ms^{-1} = 1.14 	imes 10^{-2} \ Ms^{-1}$.

The instantaneous rate of disappearance of $MnO_4^-$ ion in the following reaction is $4.56 \times 10^{-3} \ Ms^{-1}$,$2MnO_4^- + 10I^{-} + 16H^{+} \to 2Mn^{2+} + 5I_2 + 8H_2O$. The rate of appearance of $I_2$ is

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