For a chemical reaction, X → Y , the rate of reaction increases by a factor of 1.837 when the concentration of X is increased by 1.5 times, the order of the reaction with respect to X is

Question

For a chemical reaction, X → Y , the rate of reaction increases by a factor of 1.837 when the concentration of X is increased by 1.5 times, the order of the reaction with respect to X is (A) 1 (B) 1.5 (C) 2 (D) 2.5

Tardigrade Admin · Accepted Answer

Initially rate (r1)=k[X]n ldots(i) (let n be the order of reaction) then, r1×1.837=k(1.5 X)n ldots(ii) dividing (ii) by (i) 1.837=(1.5)n ∴ n=1.5 Thus, order of reaction is 1.5.

Q. For a chemical reaction, $X \to Y$ , the rate of reaction increases by a factor of $1.837$ when the concentration of $X$ is increased by $1.5$ times, the order of the reaction with respect to $X$ is

$1$

$1.5$

$2$

$2.5$

Initially rate $(r_{1}) = k [X]^{n} \dots (i)$
(let $n$ be the order of reaction)
then, $r_{1} \times 1.837 = k (1.5 X)^{n} \dots (ii)$
dividing $(ii)$ by $(i)$
$1.837 = (1.5)^{n}$
$∴ n = 1.5$
Thus, order of reaction is $1.5$ .

Q. For a chemical reaction, X→Y , the rate of reaction increases by a factor of 1.837 when the concentration of X is increased by 1.5 times, the order of the reaction with respect to X is

1

1.5

2

2.5

Initially rate (r1​)=k[X]n…(i) (let n be the order of reaction) then, r1​×1.837=k(1.5X)n…(ii) dividing (ii) by (i) 1.837=(1.5)n ∴n=1.5 Thus, order of reaction is 1.5.

Q. For a chemical reaction, $X \to Y$ , the rate of reaction increases by a factor of $1.837$ when the concentration of $X$ is increased by $1.5$ times, the order of the reaction with respect to $X$ is

$1$

$1.5$

$2$

$2.5$

Initially rate $(r_{1}) = k [X]^{n} \dots (i)$
(let $n$ be the order of reaction)
then, $r_{1} \times 1.837 = k (1.5 X)^{n} \dots (ii)$
dividing $(ii)$ by $(i)$
$1.837 = (1.5)^{n}$
$∴ n = 1.5$
Thus, order of reaction is $1.5$ .