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Question
The rate law for the reaction
\[\ce{RCl + NaOH_{(aq)} -> ROH + NaCl}\]
is given by, Rate = k [RCl]. The rate of reaction will be ______.
Options
unaffected by increasing the temperature of the reaction.
doubled on doubling the concentration of NaOH.
halved on reducing the concentration of NaOH to one half.
halved on reducing the concentration of RCl to one half.
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Solution
The rate law for the reaction
\[\ce{RCl + NaOH_{(aq)} -> ROH + NaCl}\]
is given by, Rate = k [RCl]. The rate of reaction will be halved on reducing the concentration of RCl to one half.
Explanation:
According to the rate law Rate = k [RCl], the reaction is first-order in RCl and independent of NaOH. Therefore, halving the concentration of RCl will halve the reaction rate. Changes in [NaOH] do not affect the rate.
