For the equilibrium, CH4(g) + 2 H2S(g) = CS2(g) + 4H2(g), the concentrations at equilibrium are...

Question

Question

science chemistry

Answer 1

Answer #1

CH, (g) + 2H2S 19) - (S209) + 4H219) Ke = [652] [ M]* [244] [H2S] - ci) Giveni [CH2) = 0.3322M ; [H23 )=0-6644 M; [652] = 0.0

Answer 2

Similar Homework Help Questions

QUESTION 2 CH4(8) + 2 H2S(g) = CS2(8) + 4H2(8) A reaction mixture initially contains 0.84...

QUESTION 2 CH4(8) + 2 H2S(g) = CS2(8) + 4H2(8) A reaction mixture initially contains 0.84 M CH4 and 0.45 M H2S. If the equilibrium concentration of H2 is 0.64 M, find the equilibrium constant (Ke) for the reaction *Please report 3 significant figures. Numbers only. No unit. No scientific notation
calculate the value for the thermodynamic equilibrium constant for the following reaction. CS2 (g) + 4H2...

calculate the value for the thermodynamic equilibrium constant for the following reaction. CS2 (g) + 4H2 (g) --> CH4 (g) + 2H2S (g) Values for delta Gf: CS2 (g) --> + 66.85 kJ/mol H2 (g) --> + 0.00 kJ/mol CH4 (g) --> - 50.80 kJ/mol H2S (g) --> - 33.33 kJ/mol R = 8.314 x 10^-3 kJ/molK T = 298 K
Consider the following reaction: CS2 (g) + 4 H2 (g) <====> CH4 (g) + 2 H2S...

Consider the following reaction: CS2 (g) + 4 H2 (g) <====> CH4 (g) + 2 H2S (g) Calculate the value of Keq if the equilibrium concentrations are [CS2] = 0.160, [H2] = 0.350, [CH4] = 0.0100, and [H2S] = 0.0200. a) 0.00357 b) 0.000833 c) 0.0000816 d) 600 e) 0.00167

CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.65 M...

CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.65 M CH4 and 0.82 M H2S. If the equilibrium concentration of H2 is 0.88 M, find the equilibrium constant (Kc) for the reaction. *Please report 3 significant figures. Numbers only, No unit. No scientific notation.
The initial partial pressures of H2, CS2, H2S, and CH4 in the fixed-volume reaction vessel were...

The initial partial pressures of H2, CS2, H2S, and CH4 in the fixed-volume reaction vessel were 378, 252, 68, and 54 torr, respectively. The following reaction was allowed to come to equilibrium 4H2(g) + CS2(g) ⇌ 2H2S(g) + CH4(g) Find the value of Kp given that the total pressure of the equilibrium mixture was 646 torr.
QUESTION 1 CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g) A reaction mixture initially contains...

QUESTION 1 CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g) A reaction mixture initially contains 0.55 M CH4 and 0.53 M H2S. If the equilibrium concentration of H2 is 0.48 M, find the equilibrium constant (Kc) for the reaction. *Please report 3 significant figures. Numbers only, No unit. No scientific notation.

Consider the following reaction: CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture...

Consider the following reaction: CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.800 M CH4 and 0.950 M H2S. If the equilibrium concentration of H2 is 0.440 M, find the equilibrium constant (Kc) for the reaction. A. A) 0.234 B. B) 0.0381 C. C) 2.93 D. D) 10.2 E. E) 0.0112 F. F) none of these
The equilibrium constant KC for the equation CS2 + 4H2 = CH4 + 2H2S at 900...

The equilibrium constant KC for the equation CS2 + 4H2 = CH4 + 2H2S at 900 °C is 27.8. What is the value of KC for the following equation? 1/5CH4 + 2/5 H2S = 1/5CS2 + 4/5H2
QUESTION 3 CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.65...

QUESTION 3 CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.65 M CH4 and 0.82 M H2S. If the equilibrium concentration of H2 is 0.88 M, find the equilibrium constant (Kc) for the reaction. *Please report 3 significant figures. Numbers only, No unit. No scientific notation.

Consider the following reaction: CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)A reaction mixture initially...

Consider the following reaction: CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)A reaction mixture initially contains 0.50 M CH4 and 0.75 M H2S. If the equilibrium concentration of CS2 is 0.15 M, what is the equilibrium constant (Kc) for the reaction.Hint: You might need an I.C.E. Table for this one. please show work