(CH3)3CCl(aq) + OH¯ à (CH3)3COH(aq) + Cl¯ For the reaction represented above, th
ID: 986427 • Letter: #
Question
(CH3)3CCl(aq) + OH¯ à (CH3)3COH(aq) + Cl¯ For the reaction represented above, the experimental rate law is given as follows.
Rate = k [(CH3)3CCl] If some solid sodium solid hydroxide is added to a solution that is 0.010-molar in (CH3)3CCl and 0.10-molar in NaOH, which of the following is true?
(Assume the temperature and volumes remain constant.) (A) Both the reaction rate and k increase. (B) Both the reaction rate and k decrease. (C) Both the reaction rate and k remain the same. (D) The reaction rate increases but k remains the same. (E) The reaction rate decreases but k remains the same.
Explanation / Answer
Answer – Given, reaction –
(CH3)3CCl(aq) + OH¯ ----> (CH3)3COH(aq) + Cl¯
Rate = k [(CH3)3CCl(aq)]
So the rate of reaction is depending on the (CH3)3CCl(aq) and not on the OH-, since there is rate law contain the only (CH3)3CCl(aq). So the addition of sodium hydroxide is not any affected on the rate of reaction.
So answer for this one is - (C) Both the reaction rate and k remain the same
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