CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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: 
878753_a77dae2ef0d644f08fb2a16931ef1478.png


A
0, 1/2
loader
B
1, 1
loader
C
2, 2
loader
D
3, 1
loader

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 reactant
So, $$t_{1/2}=\cfrac {C}{([A]_0)^{n-1}}$$  $$- (i)$$; $$C$$ 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image