Question

Consider the following system at equilibrium where AH° = 10.4 kJ, and K. -1.80x10-2, at 698 K. 2HI(g) 2 H2(g) + 12() When 0.24 moles of H2(g) are added to the equilibrium system at constant temperature: The value of K. The value of Qc В к The reaction must run in the forward direction to restablish equilibrium run in the reverse direction to restablish equilibrium. remain the same. It is already at equilibrium. The concentration of I, will

Answer 1

Since we know that Kc does not change with changing the concentration of either reactants or products in equilibrium mixture . This is because Kc is only dependent to temperature only and not any other factor .

While Qc i.e reaction quotient will vary as per addition or removal of species because it is calculated for each time in the reaction but Kc can only be calculated at equilibrium . So,

The Value of Kc = 1.80 x 10^-2

Value of Qc will becomes greater than Kc

because since Qc is also the ratio of product concentration to reactant so, when H2 is added at equilibrium i.e products increases it means Qc must rise .

and when Qc > Kc then

run in the reverse direction to re-establish equilibrium .

The concentration of I2 will decrease

because system shifts to the left .