For the reaction 8 H 2 S (g) - ⇀ ↽ - 8 H 2 (g) + S8 (g) 8H2S (g) ↽--⇀8H2 (g) + S8 (g) the equilibrium concentrations were found to be [ H 2 S ] = 0.250 M, [H2S]=0.250 M, [ H 2 ] = 0.600 M, [H2]=0.600 M, and [ S 8 ] = 0.750 M. [S8]=0.750 M. What is the equilibrium constant for this reaction

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Chemistry

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1 October, 07:17

For the reaction 8 H 2 S (g) - ⇀ ↽ - 8 H 2 (g) + S8 (g) 8H2S (g) ↽--⇀8H2 (g) + S8 (g) the equilibrium concentrations were found to be [ H 2 S ] = 0.250 M, [H2S]=0.250 M, [ H 2 ] = 0.600 M, [H2]=0.600 M, and [ S 8 ] = 0.750 M. [S8]=0.750 M. What is the equilibrium constant for this reaction

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Answers (1)

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Hugo Maynard · Answer 1

Kc = 826

Explanation:

For the equilibrium:

8 H₂S (g) ⇄ 8 H₂ (g) + S₈ (g)

the equilibrium constant is:

Kc = [ H₂ ] ⁸ x [ S₈ ] / [H₂S]⁸

Kc = 0.600⁸ x 0.750 / (0.250) ⁸ = 1.68 x 10⁻² x 0.750 / 1.53 x 10⁻⁵

= 824

Be careful with raising to the power of 8 in your calculator.