At a certain temperature, the equilibrium constant (Kc) for this reaction is 53.3:H2(g) + I2(g) ⇌ 2HI(g), Kc = 53.3.At this temperature, 0.700 moles of H2 and 0.700 moles of I2 were placed in a 1.00-L container to react. What is the concentration of HI at equilibrium?

To find the concentration of HI at equilibrium, we can use an ICE table (Initial, Change, Equilibrium) to keep track of the concentrations of the reactants and products.

1. Write the balanced chemical equation: H2(g) + I2(g) ⇌ 2HI(g)

2. Set up the ICE table for the initial amounts, the change, and the equilibrium amounts of the substances:

| | H2 | I2 | HI | |----------|----|-----|-----| | Initial |0.7 | 0.7 | 0 | | Change | -x | -x | +2x | | Equilibrium|0.7-x|0.7-x|2x |

3. Write the expression for Kc for this reaction: Kc = [HI]^2 / ([H2][I2]) Given Kc = 53.3

4. Plug the equilibrium concentrations into the Kc expression: 53.3 = (2x)^2 / ((0.7 - x)(0.7 - x))

5. Solve for x: 53.3 = 4x^2 / (0.49 - 1.4x + x^2) 53.3 * (0.49 - 1.4x + x^2) = 4x^2 26.117 - 74.62x + 53.3x^2 = 4x^2

Considering that Kc is large (53.3), we can assume that x will not be in the same scale as the initial amounts, which permits us to simplify the equation by neglecting the x^2 term in the denominator. This implies that nearly all of the H2 and I2 will be used to form HI.

Now solve the simplified equation: 26.117 - 74.62x ≈ 4x^2 53.3x^2 - 74.62x + 26.117 = 0

This is a quadratic equation of the form ax^2 + bx + c = 0.

6. To solve for x, we can use the quadratic formula, x = (-b ± √(b²-4ac)) / (2a). However, we'll simplify the calculation assuming the above approximation holds true:

x ≈ 26.117 / 74.62

7. Calculate x:

x ≈ 0.35

8. Determine the equilibrium concentration of HI, which is 2x:

[HI] = 2x = 2 * 0.35 = 0.70 M

Therefore, the concentration of HI at equilibrium will be approximately 0.70 M.