Question

In: Chemistry

Consider the following system at equilibrium where H° = 87.9 kJ/mol, and Kc = 1.20×10-2 ,...

Consider the following system at equilibrium where H° = 87.9 kJ/mol, and Kc = 1.20×10-2 , at 500 K. PCl5 (g) PCl3 (g) + Cl2 (g) When 0.17 moles of PCl5 (g) are removed from the equilibrium system at constant temperature: the value of Kc A. increases. B. decreases. C. remains the same. the value of Qc A. is greater than Kc. B. is equal to Kc. C. is less than Kc. the reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. the concentration of PCl3 will: A. increase. B. decrease. C. remain the same.

Solutions

Expert Solution

Sol :-

Reaction quotient (Qc) is ratio of product of the molar concentration of products to the product of the molar concentration of reactants raise to the power of stoichiometric coefficient at any stage of the reaction.

Equilibrium constant (Kc) is ratio of product of the molar concentration of products to the product of the molar concentration of reactants raise to the power of stoichiometric coefficient at equilibrium stage of the reaction.

If Qc < Kc, then reaction shift towards forward direction.

If Qc = Kc, then reaction is in equilibrium constant.

If Qc > Kc, then reaction shift towards backward direction.

Given reaction is :

PCl5 (g) <--------------------> PCl3 (g) + Cl2 (g)

When 0.17 moles of PCl5 are removed then reaction shift towards backward direction and therefore

1). Kc decreases

2). Qc greater than Kc

3). Backward direction.

4). Concentration of PCl3 decreases.

Because according to Le-Chatelier's principle when concentration of reactants decreases then equilibrium will shift towards backward direction.

queen_honey_blossom answered 2 years ago
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