Consider the following reaction: 2NH_3(g) N_2(g) + 3H_2(g) If 9.99x10^-4 moles o

Question

Consider the following reaction: 2NH_3(g) N_2(g) + 3H_2(g) If 9.99x10^-4 moles of NH_3(g), 0.224 moles of N_2, and 0.532 moles of H_2 are at equilibrium in a 11.5 L container at 937 K. the value of the equilibrium constant. K-c. is A student ran the following reaction in the laboratory at 302 K: 2NO(g) + Br_2(g) 2NOBr(g) When she introduced 0.133 moles of NO(g) and 0.119 moles of Br_2(g) into a 1.00 liter container, she found the equilibrium concentration of Br_2(g) to be 6.87x10^-2 M. Calculate the equilibrium constant, K_e., she obtained for this reaction. The equilibrium constant, K_c, for the following reaction is 1.59 at 668 K. 2NH_3(g)=N_2(g) 3H_2(g) When a sufficiently large sample of NH_3(g) is introduced into an evacuated vessel at 668 K, the equilibrium concentration of H_2(g) is found to be 0.357 M. Calculate the concentration of NH_3 in the equilibrium mixture. M The equilibrium constant, K_e, for the following reaction is 1.59 at 668 K. 2NH_3(g)=N_2(g). 3H_2 When a sufficiently lar e sample of NH3(g) is introduced into an evacuated vessel at 668 K, the equilibrium concentration of H2(g) is found to be 0.357 M. Calculate the concentration of NH3 in the equilibrium mixture. M The equilibrium constant, K,.. for the following reaction is 6.50* 10 3 at 298K 2NOBr(g)=^2NO(g) Br2(g)

Explanation / Answer

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

[NH3] = (9.99*10^-4)/11.5 = 0.00008686

[N2] = 0.224/11.5 = 0.0194

[H2] = 0.532/11.5 = 0.04626086956

then

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

Kc = (0.0194)(0.04626086956)^3 / (0.00008686)^2

Kc = 254.567

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