Chemistry · 2024
JEE Main · 4 April 2024, Shift 2 · Q63
The equilibrium constant for the reaction SO_3( g ) ⇌ SO_2( g )+1/2 O_2( g ) is K_c=4.9 × 10^-2. The value of K_c for the reaction given below is 2…
The equilibrium constant for the reaction
$$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})
$$
is $\displaystyle K_c=4.9 \times 10^{-2}$. The value of $\displaystyle K_c$ for the reaction given below is
$\displaystyle 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ is :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 416$
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.