For the reaction represented below, the experimental rate law is given by Rate =

Question

For the reaction represented below, the experimental rate law is given by Rate = k [(CH_3)_3CCl]. (CH_3)_3CCl(aq) + OH^-(aq) rightarrow (CH_3)_3COH(aq) + Cl^-(aq) If some solid sodium hydroxide were added to a solution in which (CH_3)_3CCI] = 0.01 M and [NaOH] = 0.10 M, which of the following statements would be true? (Assume the temperature and volume remain constant.) A) Both the reaction rate and k would increase. B) Both the reaction rate and k would decrease. C) Both the reaction rate and k would remain the same. D) The reaction rate would increase but k would remain the same. E) The reaction rate would decrease but k would remain the same.

Explanation / Answer

the rate of reaction , (CH3)3CCl+ OH- ------------->(CH3)3COH+Cl- , r= K[CH3)3CCl]

K =rate constnat which constant

for elementray reaction as above, r= K1[CH3)3CCl][OH-], K1 rate constant for forward reaction.

when [OH-] is very high compared to [CH3)3CCl] , the change in [OH-] is almost negligible and hence

the rate law can be written as r= K[CH3)3CCl], where K= K1[OH-], K1 the rate constant is a function of only temperature and hecne constant for the system,

[OH-] is coming from NaOH. when more NaOH is added, it will increase [OH-] concentration. So K increases. hence the rate also increases. So A is the correct ansswer.

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