Question

For the reaction 2H1 H_2(g) + I_2(g) at 698 K, the initial concentrations are [HI]: 1.000 M, and 0.000 for both products. Kc= 1.84 x 10-2 at 698 K. Write the ICE (initial, change, equilibrium) table for this reaction, and calculate the equilibrium concentrations of each reactant and product at equilibrium.

Answer 1

The equilibrium concentrations are [H₂] = -0.538 M, [I₂] = -0.269 M, and [HI] = 1.538 M. To calculate the equilibrium concentrations for the reaction 2H₂(g) + I₂(g) → 2HI(g) at 698 K, we can use the ICE table method.

Step-by-step explanation:

Constructing the ICE table for the reaction 2H₂(g) + I₂(g) → 2HI(g) at 698 K involves determining the initial, change, and equilibrium concentrations for each species.

With initial concentrations of [HI] at 1.000 M and [H₂] and [I₂] both at 0.000 M, stoichiometry and the equilibrium constant (Kc = 1.84 x 10⁻²) guide the calculation of equilibrium concentrations.

Expressing changes as -2x for [H₂], -x for [I₂], and +2x for [HI], the equilibrium concentrations become [H₂] = 0 - 2x, [I₂] = 0 - x, and [HI] = 1.000 + 2x.

Applying the equilibrium constant expression, Kc = [HI]² / ([H₂]² [I₂]), yields the quadratic equation 1.84 x 10⁻² = (1.000 + 2x)² / ((0 - 2x)² (0 - x)).

Solving for x results in x = 0.269.

Substituting this value back into the equilibrium concentrations, [H₂], [I₂], and [HI] are determined as -0.538 M, -0.269 M, and 1.538 M, respectively.

This comprehensive process provides a detailed understanding of the chemical equilibrium state at the specified temperature.

Answer 2

In this case, the equilibrium concentrations are [HI] = 0.528 M, [H₂] = 0.236 M, and [I₂] = 0.236 M.

Step-by-step explanation:

To solve this question, we will use the ICE (Initial, Change, Equilibrium) table method. We start with the initial concentrations given:

[HI] = 1.000 M, [H₂] = 0.000 M, and [I₂] = 0.000 M.

Since the reaction produces 1 mole of H₂ and I₂ for every 2 moles of HI, we can write the change in concentration for HI as -2x, and the changes for H₂ and I₂ as +x each.

At equilibrium, the equilibrium concentrations will be 1.000 - 2x for HI, and x for both H₂ and I₂.

Finally, we use the equilibrium constant expression, Kc = [H₂][I₂]/[HI]², and substitute the equilibrium concentrations to solve for x.

Plugging in the values, we have Kc = (x)(x)/(1.000 - 2x)² = 1.84 x 10⁻²

Simplifying the equation and solving for x gives a value of x = 0.236 M.

Therefore, at equilibrium, the concentrations are [HI] = 1.000 - 2(0.236) = 0.528 M, [H₂] = [I₂] = 0.236 M.