what is the equilibrium constant for the reaction H2(g)+I2(g) 2HI(g) equilibrium

Question

what is the equilibrium constant for the reaction H2(g)+I2(g) 2HI(g) equilibrium concentrations ar [HI]=1.24M [H2]=0.140M [I2]=0.140M

(2) the reaction CS2(g)+4H2(g) CH4(g)+2H2S(g) has an equilibrium constant of 0.28 at 900 celsius. If the initial concentrations are [CS2]=0.156M [H2]=1.12M [CH4]=0.998M [H2S]=1.33M, which of the following statements is true? (a) The stystem is at equlibrium and will not shift in either direction (b) the system is not at equilibrium and will shift to the left to establish equilibrium (c) the system is not at equilibrium and will shift to the right to establish equilibrium (d) cannot be predicted

(3) the decompostition of ammonia is 2NH3(g)N2(g)+3H2(g). If Kp is 1.5*10^3 at 400 celsius what is the partial pressure of ammonia at equilibrium when N2 is 0.40atm and H2 is 0.15atm?

Explanation / Answer

H2(g)+I2(g) 2HI(g)

Kc = [HI]2/[H2][I2]

Kc = 1.242/0.142

Kc = 78.45

CS2(g)+4H2(g) CH4(g)+2H2S(g)

Kc = [CH4][H2S]2/[CS2][H2]4

Q = 0.998 x 1.332/0.156 x 1.124

Q = 7.191

Since Q>K The system will shift towards product to the right

SO the correct statement is

(c) the system is not at equilibrium and will shift to the right to establish equilibrium

3.  2NH3(g)=N2(g)+3H2(g)

Kp = [H2]3[N2]/[NH3]2

1.5 x 103 = 0.153 x 0.4/[NH3]2

[NH3] = 9.48 x 10-4 atm

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