Consider the following reaction where K_c = 55.6 at 698 K. H_2(g) + I_2(g) 2HI(g
ID: 969745 • Letter: C
Question
Consider the following reaction where K_c = 55.6 at 698 K. H_2(g) + I_2(g) 2HI(g) A reaction mixture was found to contain 2.58 times 10^-2 moles of H_2(g), 4.52 times 10^-2 moles of I_2(g) and 0.302 moles of HI(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, Q_c, equals The reaction must run in the forward direction to reach equilibrium. must run in the reverse direction to reach equilibrium. is at equilibrium.Explanation / Answer
Q = [HI]^2 / [H2][I2]
Q = (0.302^2)/((2.58*10^-2)(4.52*10^-2)) = 78.2088
since
Q > K
78.2088> 55.6
this reation is NOT in equilbirium
and will shift toward PRODUCT formation