The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD (g) ⇌ H2 (g) + D2 (g) Kc = 0.28 5 H2 (g) + 5 D2 (g) ⇌ 10 HD (g) Kc = ? The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD (g) ⇌ H2 (g) + D2 (g) Kc = 0.28 5 H2 (g) + 5 D2 (g) ⇌ 10 HD (g) Kc = ? 581 0.73 1.3 0.0017 2.06

Question

Chemistry

Pineda

8 March, 12:43

The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD (g) ⇌ H2 (g) + D2 (g) Kc = 0.28 5 H2 (g) + 5 D2 (g) ⇌ 10 HD (g) Kc = ? The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD (g) ⇌ H2 (g) + D2 (g) Kc = 0.28 5 H2 (g) + 5 D2 (g) ⇌ 10 HD (g) Kc = ? 581 0.73 1.3 0.0017 2.06

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Cason Casey · Answer 1

Answer choice 'D' = > Kc (2) = 581

Explanation:

2HD ⇄ H₂ + D₂ = > Kc (1) = [H₂][D₂]/[HD]² = > 0.28

5H₂ + 5D₂ ⇄ 10HD = > Kc (2) = [HD]¹⁰/[H₂]⁵[D₂]⁵ = 1 / (Kc (1)) ⁵ = 1 / (0.28) ⁵ = 581

For these type problems, one should 1st write the empirical Kc expression for each equation given. Then compare the expressions and ask 'how can the 1st Kc expression be changed into the 2nd Kc expression?' Apply and substitute given Kc (1) into the Kc (2) expression and solve for numerical results.