A mixture containing 3.7 moles of NO and 0.88 mole of CO2 was allowed to react i

Question

A mixture containing 3.7 moles of NO and 0.88 mole of CO2 was allowed to react in a flask at a certain temp according to the equation....

NO + CO2 <---> NO2 + CO

At equilibrium, 0.15 mole of CO2 was present. Calculate the equilibrium constant Kc of this reaction.

Explanation / Answer

Lets ignore the chemicals and rewrite the equation as follows: A + B C + D The forward reaction rate of this equilibrium reaction is: Rf = kf*nA*nB*V^2 The backward reaction rate of this equilibrium reaction is: Rb = kb*nC*nD*V^2 where nA-nD denote the amount of moles of species A-D, V is the volume of the container, k is the reaction constant and f and b denote forward and backward respectively. At equilibrium the forward reaction goes as fast as the backward reaction so Rf = Rb -> kf*nA*nB*v^2 = kb*nC*nD*V^2 defining the equilibrium constant Kc = kf/kb = (nC*nD)/(nA*nB) before the reaction: nA = 3.7 mol, nB = 0.88 mol, nC = 0 mol, nD = 0 mol after reaction: nA = x mol, nB = 0.15 mol, nC = y mol, nD = z mol difference: dA = 3.7-x mol, dB = 0.77 mol, dC = -y mol, dD = -z mol where x, y, z are the amount of moles of species A,C and D that are needed to calculate the equilibrium constant.According to the reaction for every mole of A converted, one mole of B is converted and one mole of C and D is produced. This implies that -dA = -dB = dC = dD, from this follows that x = 3.13 mol and y = z = 0.77 mol. Substituting into the formula for the equilibrium constant yields: Kc = (0.77*0.77)/(3.13*0.11) = 1.722

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