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.

A) For the reaction

3A(g)+2B(g)⇌C(g)

Kc = 50.6 at a temperature of 355 ∘C .Calculate the value of Kp.

B) For the reaction

X(g)+3Y(g)⇌3Z(g)

Kp = 1.76×10−2 at a temperature of 273 ∘C. Calculate the value of Kc.

A)

3A(g)+2B(g)⇌C(g), Kc = 50.6

Δn = (Sum of Number of moles of Products) - (Sum of Number of moles of Reactants)/

Δn = 1 - ( 3 + 2) = 1 - 5 = -4

and T = 3550C

T = 355+ 273 = 656 K and R= 0.08206L

Relation between Kc and Kp is, Kp = Kc (Rx T) Δn

so Kp = 50.6 X ( 0.08206 X 656 ) -4 = 2723.86X 10-4

Kp = 2723.86X 10-4

B)

X(g)+3Y(g)⇌3Z(g)

Kp = 1.76×10−2 at T = 273 C, R= 0.08206 L

T = 273+ 273 = 546K

Δn = 3 - ( 1 +3) = 3-4 = -1

Kp = Kc (Rx T) Δn

Kc = Kp /(Rx T) Δn

Kc = 1.76×10−2/(0.08206×546)-1

=1.76×10−2×44.80476

Kc = 78.856×10−2

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