At 500 K, the equilibrium constant for the dissociation reaction H2(g) ⇌ 2H(g) is very small (Kc = 1.2 × 10⁻⁴²). (a) What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.10 M? (b) How many H atoms and H2 molecules are present in 1.0 L of 0.10 M H2 at 500 K?
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Step 1: Write the expression for the equilibrium constant (Kc) for the reaction H2(g) ⇌ 2H(g). The expression is Kc = [H]^2 / [H2].
Step 2: Substitute the given values into the Kc expression. You know Kc = 1.2 × 10⁻⁴² and [H2] = 0.10 M. So, 1.2 × 10⁻⁴² = [H]^2 / 0.10.
Step 3: Solve for [H] by rearranging the equation to find [H]^2 = Kc × [H2]. Then, calculate [H] by taking the square root of the result.
Step 4: For part (b), calculate the number of moles of H2 in 1.0 L of 0.10 M solution using the formula: moles = concentration × volume.
Step 5: Use the concentration of H atoms found in part (a) to calculate the number of moles of H atoms in 1.0 L, and then convert these moles to the number of atoms using Avogadro's number.