Chem 152 Assignment- Chemical Equ Due Name: You need to show your work, includin

Question

Chem 152 Assignment- Chemical Equ Due Name: You need to show your work, including units, and results with the proper number of significant figures in order to receive full credit. 1. Reaction mixture contains o.417 mol/L N2.0.524 mol/L Ha, and 0.122 mo/L NHs at 800. K. At this temperature K 0.278 for the reaction: N2(g) 3 H: (g) 2 NH3 (g) a) Calculate the reaction quotient Q? b) Is there a reaction mixture at equilibrium? Explain. c) If not, is there a tendency to form more reactants or more products? Page 1 of 4

Explanation / Answer

N2 (g) + 3 H2 (g) -----> 2 NH3 (g)
Q = [NH3]^2 / [N2] [H2]^3
Q = (0.122)^2 / ((0.417)*(0.524)^3)
Q = 0.2481
Q < Kc so the reaction tend to form products.since the concentration of reactants is high.So equlibrium shifts to right

                     CH4 (g) + 2 H2S (g) ------> CS2 (g) + 4 H2 (g)
initially           0.5        0.75            0          0
at equilib (0.5 - x) (0.75 -2x)         x          4x
but 4x = 0.44 then x = 0.11
Kc = [CS2][H2]^4 / [CH4][H2S]^2
Kc = ((0.11)*(0.44)^4)/((0.5-0.11)*(0.75-0.22)^2)
Kc = 0.03763

HCl (g) + I2 (s) ------> 2 HI (g) + Cl2 (g)
Kc = [HI]^2[Cl2] / [HCl][I2]
Kc = (5.6 * 10^-16)^2(0.0019) / (0.13)(1)
Kc = 4.583 * 10^-33

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