HW2_11 - [A] [B] [C] initial 1.0 1.4 change final 0.6 The...

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Homework due 2/11 Five moles of A and 7 moles of B are placed in a 5 liter container. If at equilibrium, 3 moles of C are present, calculate K c for the equilibrium 3A + B 2 C. In order to determine the value of K c , the concentration of all species present at equilibrium needs to be known. Since the initial concentration of all species is given with [C] = 0, one can determine the concentrations at equilibrium by measuring the change in the number of moles of C, and correcting for the stoichiometry of the reaction. Based on the values given in the problem, the following concentrations can be calculated. Initial [A] = 5 moles A/5 liters = 1.0 M Initial [B] = 7 moles B/5 liters = 1.4 M Final [C] = 3 moles C/5 liters = 0.6 M Putting this in table form gives:
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Unformatted text preview: [A] [B] [C] initial 1.0 1.4 change final 0.6 The concentration of C changes from 0 M to 0.6 M; hence it increases by 0.6 M. The concentration of B will decrease by half this amount since 1 mole of B produces 2 moles of C, thus change in [B] = 0.5(0.6) = 0.3. It would be negative, since B is being used up in the process. The change in the[A] is equal to 3/2 the change in the [C], since it takes 3 moles of A to produce 2 moles of C. Thus, the change in [A] = 3(0.6)/2 = 0.9. Again this would be negative since A is being consumed in the reaction. Completing the table gives: [A] [B] [C] initial 1.0 1.4 change-0.9-0.3 +0.6 final 1.0 - 0.9 = 0.1 1.4 - 0.3 = 1.1 0 + 0.6 = 0.6 K c = [C] 2 /[A] 3 [B] = (0.6) 2 /(0.1) 3 (1.1) = 327...
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