MCQ
Conc. $H_2SO_4$ cannot be used to prepare $HBr$ from $NaBr$ because it
- Areacts slowly with $NaBr$
- Boxidises $HBr$
- Creduces $HBr$
- ✓disproportionates $HBr$
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${H_2}C = C{H_2}(g) + {H_2}(g) \to {H_3}C - C{H_3}(g)$ at $298 \,K$ will be ....$kJ$
