Following is the graph between log t1 / 2 and loga ( texta = Concentration of reactant) for a given reaction at 27° C . Hence, order of the reaction is (Here, t1 / 2 is half-life)

Question

Following is the graph between log t1 / 2 and loga ( texta = Concentration of reactant) for a given reaction at 27° C . Hence, order of the reaction is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-fzo93pgxfytekbwm.jpg" /> (Here, t1 / 2 is half-life) (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

t (1/2) propto((1/ a )) n -1 for nth order reaction. beginarrayl t (1/2)= k ( a )1- n log t (1/2)=(1- n ) log a + log k endarray Thus, graph between log t (1/2) and log a is linear slope. (1-n)= tan 45° n =0

Q. Following is the graph between $l o g t_{1/2}$ and $l o g a$ ( $a =$ Concentration of reactant) for a given reaction at $2 7^{\circ} C$ . Hence, order of the reaction is

(Here, $t_{1/2}$ is half-life)

$0$

$1$

$2$

$3$

$t_{\frac{1}{2}} \propto (\frac{1}{a})^{n - 1}$ for nth order reaction.
$t_{\frac{1}{2}} = k (a)^{1 - n} lo g t_{\frac{1}{2}} = (1 - n) lo g a + lo g k$
Thus, graph between $lo g t_{\frac{1}{2}}$ and $lo g$ a is linear slope.
$(1 - n) = tan 4 5^{\circ}$
$n = 0$

Q. Following is the graph between logt1/2​ and loga ( a= Concentration of reactant) for a given reaction at 27∘C . Hence, order of the reaction is (Here, t1/2​ is half-life)

0

1

2

3

t21​​∝(a1​)n−1 for nth order reaction. t21​​=k(a)1−nlogt21​​=(1−n)loga+logk​ Thus, graph between logt21​​ and log a is linear slope. (1−n)=tan45∘ n=0

Q. Following is the graph between $l o g t_{1/2}$ and $l o g a$ ( $a =$ Concentration of reactant) for a given reaction at $2 7^{\circ} C$ . Hence, order of the reaction is

(Here, $t_{1/2}$ is half-life)

$0$

$1$

$2$

$3$

$t_{\frac{1}{2}} \propto (\frac{1}{a})^{n - 1}$ for nth order reaction.
$t_{\frac{1}{2}} = k (a)^{1 - n} lo g t_{\frac{1}{2}} = (1 - n) lo g a + lo g k$
Thus, graph between $lo g t_{\frac{1}{2}}$ and $lo g$ a is linear slope.
$(1 - n) = tan 4 5^{\circ}$
$n = 0$