For a chemical reaction $A+B\rightarrow$ Product, the order is $1$ with respect to $A$ and $B$. Rate $\mathrm{mol}{L}^{-1}{S}^{-1}$ [A] $\mathrm{mol}{L}^{-1}$ [B] $\mathrm{mol}{L}^{-1}$ $0.10$ $20$ $0.5$ $0.40$ $x$ $0.5$ $0.80$ $40$ $y$ What is the value of $x$ and $y$?

JEE Main 2023 — Chemistry Physical Chemistry

2023mcqeasy

For a chemical reaction $A+B\rightarrow$ Product, the order is $1$ with respect to $A$ and $B$ .

Rate $\mathrm{mol}{L}^{-1}{S}^{-1}$	[A] $\mathrm{mol}{L}^{-1}$	[B] $\mathrm{mol}{L}^{-1}$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

What is the value of

x

and

y

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

Options

A
$160$ and $4$
B
$80$ and $4$
C
$80$ and $2$
D
$40$ and $4$

Did you get this right?

Solution

To determine the values of x and y, we can use the information provided and the rate expression for the given reaction, which is first order with respect to both A and B:

Rate $r=K[A{]}^{1}[B{]}^{1}$

$0.1=K(20{)}^{1}(0.5{)}^{1}\ldots (i) 0.40=K(x{)}^{1}(0.5{)}^{1}\ldots (\mathrm{ii}) 0.80=K(40{)}^{1}(y{)}^{1}\ldots (\mathrm{iii})$

From (i) and (ii)

$x=80$

From (i) and (iii)

$y=2$

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Solution

To determine the values of x and y, we can use the information provided and the rate expression for the given reaction, which is first order with respect to both A and B:

Rate $r=K[A{]}^{1}[B{]}^{1}$

$0.1=K(20{)}^{1}(0.5{)}^{1}\ldots (i) 0.40=K(x{)}^{1}(0.5{)}^{1}\ldots (\mathrm{ii}) 0.80=K(40{)}^{1}(y{)}^{1}\ldots (\mathrm{iii})$

From (i) and (ii)

$x=80$

From (i) and (iii)

$y=2$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main 2023 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry