JEE Main 2025 Jan 23 Shift 2, Chemistry Q20: Factors Affecting Equilibria

JEE Main2025Jan 23, Shift 2Chemistry

Consider the following reactions

$K_{2} C r_{2} O_{7} K O H - H_{2} O [A] H_{2} S O_{4} - H_{2} O [B] + K_{2} S O_{4}$

The products [A] and [B], respectively are :

Solution

The chromate $-$ dichromate interconversion depends on pH.

In alkaline solution (KOH) the orange dichromate is converted to yellow chromate:

$C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ , so [A] $= K_{2} C r O_{4}$ .

Acidifying chromate with H $_{2}$ SO $_{4}$ reverses the process and regenerates orange dichromate:

$2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ , so [B] $= K_{2} C r_{2} O_{7}$ (with K $_{2}$ SO $_{4}$ as a by-product).

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.