Consider the following reaction where K 6.30 at 723 K 2NH3(g) N2g) 3H2(g) A reac

Question

Consider the following reaction where K 6.30 at 723 K 2NH3(g) N2g) 3H2(g) A reaction mixture was found to contain 6.03x10-4 moles ofNH3(g), 2.33-10* moles of N2(g), and 4.17x102 moles of H2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qe equals The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Explanation / Answer

Q = [N2][H2]^3 /[NH3]^2

Q = (2.33*10^-2)((4.17*10^-2)^3) / ((6.03*10^-4)^2)

Q = 4.6465

since Q > K

there are more products than reactants

then

products must form reactants

choose B

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