Ans. Balanced reaction: H2(g) + I2(g) ------------> 2 HI(g)
Equilibrium constant, Kc = [HI]2 / ([H2] [I2]) - equation 1
Part A: Putting the values in equation 1-
Kc = (0.922)2/ (0.0355 x0.0388) = 0.850084 / 0.0013774 = 617.16
Hence, Kc = 617
Part B. Putting the values in equation 1-
90.6 = (0.385)2/ ([H2] x 4.6 x10-2)
Or, 90.6 x 0.046 [H2] = 0.148225
Or, [H2] = 0.148225 / 4.1676 = 0.04
Thus, [H2] = 0.04 M
Part C: Putting the values in equation 1-
50.2 = ([HI])2/ (4.9 x 10-2 x 4.64 x10-2)
Or, ([HI])2 = 50.2 x 0.0022736 = 0.11413472
Or, [HI] = (0.11413472)1/2 or under root (0.11413472)
Hence, [HI] = 0.34
Thus, [HI] = 0.34
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