For the following reaction, K_p = 137 at 3.50 $\times$10²K. What is the value of K_c? 2C₄H₁₀(l) + 3O₂(g) 8CO₂(g) + 10H₂O(g)

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Using the equation K_p = 137 and the stoichiometric coefficients given in the balanced chemical equation, we can set up an ICE (initial, change, equilibrium) table for the reaction: | | 2C₄H₁₀(l) | F1F1F1S1F1F1F10 | 3O₂(g) | 8CO₂(g) | 10H₂O(g) | |----------|--------------------------------|------------------|-------------------|---------------------|---------------------| | Initial | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | | Change | -x | -2x | 3x | 8x | 10x | | Equilibrium | 0.01 - x | 0.01 - 2x | 0.01 + 3x | 0.01 + 8x | 0.01 + 10x | where x is the change in concentration or pressure for each species. We can plug the equilibrium concentrations or pressures into the equilibrium expression for the reaction: K_c = [F1F1F1S1] / ([C₄]2[H₁₀][O₂]3/[CO₂]8/[H₂O]10) K_c = (0.01 - 2x) / (0.01)2(0.01 + 3x)3 / (0.01 + 8x)8 / (0.01 + 10x)10 Using algebra, we can simplify and solve for x: 137 = K_p = K_c 137 = (0.01 - 2x) / (0.01)2(0.01 + 3x)3 / (0.01 + 8x)8 / (0.01 + 10x)10 x = 7.03x10-9 Substituting x back into the equilibrium concentrations or pressures gives us: [C₄] = 0.01 - x = 0.01 - 7.03x10-9 = 0.01 [F1F1F1S1] = 0.01 - 2x = 0.01 - 2(7.03x10-9) = 0.00999998594 [H₁₀] = 0.01 [O₂] = 0.01 + 3x = 0.01000002109 [CO₂] = 0.01 + 8x = 0.01000005626 [H₂O] = 0.01 + 10x = 0.01000007029 Therefore, the equilibrium constant K_c is approximately 7.03x1010, which corresponds to choice A.

For the Following Reaction, Kp = 137 at 3 ×\times× 102K ×\times×

For the Following Reaction, K_p = 137 at 3 $\times$ 10²K $\times$