Question

# The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the...

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation

Kp=Kc(RTn

where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction

N2(g)+3H2(g)⇌2NH3(g)

for which Δn=2−(1+3)=−2.

For the reaction

2A(g)+2B(g)⇌C(g)

Kc = 71.6 at a temperature of 159 ∘C .

Calculate the value of Kp.

For the reaction

X(g)+3Y(g)⇌3Z(g)

Kp = 1.82×10−2 at a temperature of 31 ∘C .

Calculate the value of Kc.

2H2S(g) ⇌ 2H2(g) + S2(g)

initiall:   3.70×10−4 0 0

change: -2x +2x +x

equalibrium: 3.70×10−4 -2x 2x x

Kc = [H2]2[S2] /[H2S]2

1.67×10−7 = (2x2 . x ) /  (3.70×10−4 - 2x)2

2x3 = 1.67×10−7 x( 3.70×10−4)2

2x3= 1.67 x 10-7 x 13.69 x 10-8

x3 =22.8623 x 10-15 / 2 = 11.43115x 10-15

x =2.253 x 10-5

At equalibrium concentration of :

[H2S] = 3.70×10−4 - 2x

= 3.70×10−4 - ( 2 x 2.25 x 10-5)

= 3.70×10−4 - 0.45 x 10-4 = 3.45 x 10-4

[H2] = 2x

= 2 x 2.253 x 10-5 = 4.506 x 10-5

[S2] = x

= 2.253 x 10-5

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