Question

# What will be the order of reaction and rate constant for a chemical change having $$\log t_{50}$$% $$vs$$ $$\log$$ concentration of $$(A)$$ curves as:

A
0, 1/2
B
1, 1
C
2, 2
D
3, 1

Solution

## The correct option is A $$0,\ 1/2$$We know that,Half time of $$n^{th}$$ order reaction $$(t_{1/2}) \propto \left( \cfrac {1}{[A]_0}\right)^{n-1}$$where $$[A]_0$$ is the initial concentration of the reactantSo, $$t_{1/2}=\cfrac {C}{([A]_0)^{n-1}}$$  $$- (i)$$; $$C$$ is proportionality constant.Taking $$\log$$ on both sides of $$(i)$$:-$$\log t_{1/2}=\log C (n-1) \log [A]$$                                 $$[\because \log \left(\cfrac {a}{b}\right)=\log a- \log b$$ & $$\log (a^b)=b\log a]$$Now, for $$n=0$$,$$\log t_{1/2}=\log C- (0-1) log [A]_0$$$$\Rightarrow log t_{1/2}=\log C+log [A]_0$$If $$C=)$$, then,$$\Rightarrow \log t_{1/2}=\log [A]_0$$          $$- (ii)$$$$Eq^n (ii)$$ represents the graph between $$\log t_{1/2}$$ & $$\log [A]_0$$ as the straight line passing through origin with an angle of $$45^\circ$$ from $$\log [A]_0$$ axis.(Similarly as line $$y=x$$)Now, for a zero order reaction,$$t_{1/2}= \cfrac {[A]_0}{2K}$$where, $$K$$ is the rate constant.$$\Rightarrow K= \cfrac {[A]_0}{2t_{1/2}}$$     $$- (iii)$$From the graph we have:-$$\log t_{1/2}=\log [A]_0$$$$\Rightarrow t_{1/2}=[A]_0$$       (Taking antilog on both sides)So, from $$(iii)$$, $$K=\cfrac {[A]_0}{2[A]_0}=\cfrac {1}{2}$$Chemistry

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