1
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For $\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$

$\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively

If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Which of the following statement is correct?

A
Catalyst does not alter the Gibbs energy change of a reaction.
B
The enthalpy change for the reaction is $+20 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
C
Catalyst can cause non-spontaneous reactions to occur.
D
The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.
2
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Predict the major product of the following reaction sequence:-

JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Hydrocarbons Question 1 English

A
JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Hydrocarbons Question 1 English Option 1
B
JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Hydrocarbons Question 1 English Option 2
C
JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Hydrocarbons Question 1 English Option 3
D
JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Hydrocarbons Question 1 English Option 4
3
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Which one of the following about an electron occupying the 1 s orbital in a hydrogen atom is incorrect?

(Bohr's radius is represented by $\mathrm{a}_0$)

A
The probability density of finding the electron is maximum at the nucleus
B
The total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus
C
The electron can be found at a distance $2 a_0$ from the nucleus
D
The 1 s orbital is spherically symmetrical
4
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Rate law for a reaction between $A$ and $B$ is given by

$$\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$$

If concentration of $A$ is doubled and concentration of $B$ is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is

A
$(\mathrm{n}-\mathrm{m})$
B
$2^{(\mathrm{n}-m)}$
C
$\frac{1}{2^{m+n}}$
D
$(\mathrm{m}+\mathrm{n})$
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