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OpenStudy (anonymous):
The value of the equilibrium constant for the following reaction is 345. A + 2B 3C + D What is the value of the equilibrium constant for the reaction below? 2A + 4B 6C + 2D a. K = (345)2 = 1.19 x 10 + 5 b. K = 2 (345)2 = 2.38 x 10 +5 c. K = (2 345)2 = 4.76 x 10 +5 d. K = (345)1/2 = 18.6 e. K = 345
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