Calculate Ke for the following - I am so lost on these..?

Question

1, Calculate Ke for the following (equilibrium concentrations given below substances)

a) H2 (g) + Br2 (g) ↔ 2 HBr (g) + heat

[0.513] [0.62] [0.346]

b) N2O4 (g) + heat ↔ 2 NO2 (g)

[0.0325] [0.022]

c) NH4SH (g) ↔ NH3 (g) + H2S (g) + heat

[0.027] [0.0234] [0.209]

Answer 1

a)

Ke = [HBr]^2 / [H2][Br2] = 0.346^2 / (0.513 x 0.62)

Ke = 0.376

Ke = product (HBr) raised to the power of the coefficient (2) divided by the reactants raised to the power of their coefficients (H2^1 and Br2^1).  

Answer 2

Didn't I answer these last night for Jennifer? Anyways,

Kc = [product c]^#[product a]^# / [reactant a]^#[reactant b]^#

Substitute in the concentrations to this equation.The power each concentration is raised to it equivalent to coefficient of the corresponding species in the equation.

a) Kc = [0.346]^2 / [0.513][0.62] = 0.38

b) Kc = [0.022]^2 / [0.0325] = 0.015

c) Kc = [0.0234][0.209] / [0.027] = 0.18

Answer 3

Just remember that Kc = [products] / [reactants]

a) Kc =  [HBr]² / ([H2][Br2]) = 0.346² / (0.513*0.62) = answer

b) Kc = [NO2]² / [N2O4] = (0.022)² / (0.0325) = answer

c) Kc = ([H2S][NH3]) / [NH4SH] = [(0.209)(0.0234)] / (0.027) = answer

Remember the coefficient of the moles becomes an exponent in the equilibrium constant, Kc/Keq.

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