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For the gas phase reaction, PCl5(g)PCl3(g)+Cl2(g)\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}, which of the following conditions is correct?

A

ΔH=0\Delta H = 0 and ΔS<0\Delta S < 0

B

ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0

C

ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0

D

ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0

Step-by-Step Solution

For the given decomposition reaction: PCl5(g)PCl3(g)+Cl2(g)\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}

  1. Enthalpy change (ΔH\Delta H): Decomposition reactions typically involve the breaking of bonds, which requires an input of energy. Therefore, the reaction is endothermic, meaning ΔH>0\Delta H > 0 (specifically, ΔrH=124.0 kJ mol1\Delta_r H^\circ = 124.0 \text{ kJ mol}^{-1}).
  2. Entropy change (ΔS\Delta S): In this reaction, 1 mole1 \text{ mole} of a gaseous reactant decomposes to form 2 moles2 \text{ moles} of gaseous products (Δng=1+11=+1\Delta n_g = 1 + 1 - 1 = +1). Since the number of gaseous particles increases, the randomness or disorder of the system also increases. Therefore, ΔS>0\Delta S > 0.
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