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2 September, 19:14

Calculate the value of the equilibrium constant, Kc, for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2, 0.0380 moles of N2 and 0.0276 moles of H2O vapor were present in a 2.00 L reaction vessel at equilibrium. 2NO (g) + 2H2 (g) N2 (g) + 2H2O (g)

6.54

0.352

3.27

0.176

+1
Answers (1)
  1. 2 September, 20:55
    0
    the reaction is

    2NO (g) + 2H2 (g) N2 (g) + 2H2O (g)

    Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

    Given

    moles of NO = 0.124 therefore [NO] = moles / volume = 0.124 / 2 = 0.062

    moles of H2 = 0.0240, therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

    moles of N2 = 0.0380, therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

    moles of H2O = 0.0276, therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

    Kc = (0.019) (0.0138) ^2 / (0.062) ^2 (0.012) ^2 = 6.54
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