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.

For the reaction

2A(g)+2B(g)⇌C(g)

*K*c = 71.6 at a temperature of 159 ∘C .

Calculate the value of *K*p.

For the reaction

X(g)+3Y(g)⇌3Z(g)

*K*p = 1.82×10^{−2} at a temperature of 31 ∘C
.

Calculate the value of *K*c.

Answer #1

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

K_{c} = [H2]^{2}[S2] /[H2S]^{2}

1.67×10^{−7 =} (2x^{2} . x )
/ (3.70×10^{−4} - 2x)^{2}

2x^{3} = 1.67×10^{−7} x(
3.70×10−4)^{2}

2x^{3}= 1.67 x 10^{-7} x 13.69 x
10^{-8}

x^{3} =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}**

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
3A(g)+2B(g)⇌C(g)
Kc =...

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 =...

For chemical reactions where all reactants and products are in
the gas phase the amount of each gas in the vessel can be expressed
either as partial pressures or as concentrations. As such the
equilibrium constant for a gas phase reaction can also be expressed
in terms of concentrations or pressures. For the general
reaction,
aA(g)+bB(g)⇌cC(g)+dD(g)
Kp=(PC)c(PD)d(PA)a(PB)b
and
Kc=[C]c[D]d[A]a[B]b.
It is possible to interconvert between Kp and
Kcusing
Kp=Kc(RT)Δn
where R=0.08314 L bar mol−1 K−1 and Δn is the
difference...

At equilibrium, the concentrations of reactants and products can
be predicted using the equilibrium constant, Kc, which is
a mathematical expression based on the chemical equation. For
example, in the reaction
aA+bB⇌cC+dD
where a, b, c, and d are the
stoichiometric coefficients, the equilibrium constant is
where [A], [B], [C], and [D] are the equilibrium concentrations.
If the reaction is not at equilibrium, the quantity can still be
calculated, but it is called the reaction quotient, Qc,
instead of the...

At equilibrium, the concentrations of reactants and products can
be predicted using the equilibrium constant, Kc, which is
a mathematical expression based on the chemical equation. For
example, in the reaction
aA+bB⇌cC+dD
where a, b, c, and d are the
stoichiometric coefficients, the equilibrium constant is
Kc=[C]c[D]d[A]a[B]b
where [A], [B], [C], and [D] are the equilibrium concentrations.
If the reaction is not at equilibrium, the quantity can still be
calculated, but it is called the reaction quotient, Qc,
instead of...

Part A
For the reaction
2A(g)+2B(g)⇌C(g)
Kc = 62.6 at a temperature of 187 ∘C .
Calculate the value of Kp.
Express your answer numerically.
---------------------------------
Part B
For the reaction
X(g)+3Y(g)⇌3Z(g)
Kp = 3.10×10−2 at a temperature of 325 ∘C
.
Calculate the value of Kc.
Express your answer numerically.

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.
Part A
For the reaction
2CH4(g)⇌C2H2(g)+3H2(g)
K = 0.155 at 1635 ∘C . What is Kp for the
reaction at this temperature?
Express...

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.'
Part A
For the reaction
2CH4(g)⇌C2H2(g)+3H2(g)
K = 0.130 at 1668 ∘C . What is Kp for the
reaction at this temperature?
Express...

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.
Part A For the reaction 2CH4(g)⇌C2H2(g)+3H2(g) K = 0.165 at 1521
∘C . What is Kp for the reaction at this temperature? Express...

1.) The equilibrium constant for the chemical equation
N2(g)+3H2(g) <-->
2NH3(g)
is Kp = 1.09 at 209 °C. Calculate the value of the Kc for the
reaction at 209 °C.
2.) At a certain temperature, 0.3411 mol of N2 and 1.581 mol of
H2 are placed in a 1.50-L container.
N2(g)+3H2(g) <-->
2NH3(g)
At equilibrium, 0.1801 mol of N2 is present. Calculate the
equilibrium constant, Kc.
3.) At a certain temperature, the Kp for the decomposition of
H2S is 0.748....

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