JEE MAIN - Chemistry (2007 - No. 25)
Consider the reaction, 2A + B $$\to$$ products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A
alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is
L mol-1 s-1
no unit
mol L-1 s-1
s-1
Explanation
Rate = k [A]x[B]y
When [B] is doubled, keeping [A] constant half-life of the reaction does not change.
For a first order reaction $${t_{1/2}} = {{0.693} \over K}\,\,$$ i.e. for a first order reaction $${t_{1/2}}$$ does not depend up on the concentration. Hence the reaction is first order with respect to B. Now when [A] is doubled, keeping [B] constant, the rate also doubles. Hence the reaction is first order with respect to A.
$$\therefore$$ $$\,\,\,\,\,$$ Rate = k [A]1[B]1
Order of reaction $$=1+1=2$$
Now for a nth order reaction, unit of rate constant is
(L)n–1 (mol)1–n s–1 when n = 2, unit of rate constant
is $$L\,\,mo{l^{ - 1}}\,{\sec ^{ - 1}}.$$
When [B] is doubled, keeping [A] constant half-life of the reaction does not change.
For a first order reaction $${t_{1/2}} = {{0.693} \over K}\,\,$$ i.e. for a first order reaction $${t_{1/2}}$$ does not depend up on the concentration. Hence the reaction is first order with respect to B. Now when [A] is doubled, keeping [B] constant, the rate also doubles. Hence the reaction is first order with respect to A.
$$\therefore$$ $$\,\,\,\,\,$$ Rate = k [A]1[B]1
Order of reaction $$=1+1=2$$
Now for a nth order reaction, unit of rate constant is
(L)n–1 (mol)1–n s–1 when n = 2, unit of rate constant
is $$L\,\,mo{l^{ - 1}}\,{\sec ^{ - 1}}.$$
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