The rate law for the reaction, $R C l + N a O H (a q .) ⟶ R O H + N a C l$ , is given by, Rate $= k [R C l]$ . The rate of reaction will be:

unaffected by increasing temperature of the reaction

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doubled on doubling the concentration of

N a O H

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halved on reducing the concentration of

N a O H

to one half

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halved on reducing the concentration of

R C l

to one half

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Solution

The correct option is D halved on reducing the concentration of $R C l$ to one half
Rate $= K [R C l]$
Hence rate will be independent of the concentration of all other reagents in the reaction and it will be proportional to the concentration of $R C l$ .
Thus, option D is correct.

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Similar questions

The rate law for the reaction:
$R C l + N a O H \to R O H + N a C l$ is given by;
Rate = $k [R C l]$ . The rate of this reaction is

(IIT-JEE, 1990)

R C l + N a O H (a q) \to R O H + N a C l

\to

=

R C l + N a O H (a q) \to R O H + N a C l

R a t e = k [R C l]

R C l + N a O H (a q) \to R O H + N a C l

R a t e = k [R C l]

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