Question

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

*K*p=*K*c(*R**T*)Δ*n*

where *R*=0.08206 L⋅atm/(K⋅mol), *T* is the
absolute temperature, and Δ*n* 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 Δ*n*=2−(1+3)=−2.

A) For the reaction

3A(g)+2B(g)⇌C(g)

*K*c = 50.6 at a temperature of 355 ∘C .Calculate the
value of *K*p.

B) For the reaction

X(g)+3Y(g)⇌3Z(g)

*K*p = 1.76×10^{−2} at a temperature of 273 ∘C.
Calculate the value of *K*c.

Answer #1

A)

3A(g)+2B(g)⇌C(g), K_{c} = 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 = 355^{0}C

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

Relation between Kc and Kp is, K_{p} = K_{c} (Rx
T) ^{Δn}

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

**K _{p} = 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

K_{p} = K_{c} (Rx T) ^{Δn}

K_{c} = K_{p}
/(Rx T) ^{Δn}

K_{c} =
1.76×10^{−2}/(0.08206×546)^{-1}

=1.76×10^{−2}×44.80476

**K _{c} = 78.856×10^{−2}**

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(RT)Δn
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...

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(RT)Δn 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 X(g)+3Y(g)⇌2Z(g) Kp =...

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either as partial pressures or as concentrations. As such the
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in terms of concentrations or pressures. For the general
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aA(g)+bB(g)⇌cC(g)+dD(g)
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and
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If the reaction is not at equilibrium, the quantity can still be
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constant K can be expressed either in terms of the concentrations
of the gases (in M) or as a function of the partial pressures of
the gases (in atmospheres). In the latter case, the equilibrium
constant is denoted as Kp to distinguish it from the
concentration-based equilibrium constant K.'
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For chemical reactions involving ideal gases, the equilibrium
constant K can be expressed either in terms of the concentrations
of the gases (in M) or as a function of the partial pressures of
the gases (in atmospheres). In the latter case, the equilibrium
constant is denoted as Kp to distinguish it from the
concentration-based equilibrium constant K.
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a mathematical expression based on the chemical equation. For
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aA+bB⇌cC+dD
where a, b, c, and d are the
stoichiometric coefficients, the equilibrium constant is
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