The equilibrium constant, K c, is calculated using molar concentrations. For gas

Question

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)n

where R=0.08206 Latm/(Kmol), T is the absolute temperature, and 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 n=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 = 1.82×102 at a temperature of 31 C .

Calculate the value of Kc.

HEM 136 Winter 2017 Signed in as Amal Hage Heu H15 HLN t Pressure-Based versus Concentration-Based Equilibrium Constants previous l 40 of5 Pressure-Based versus Concentration- Based Equilibrium Constants Part A For the reaction The equilibrium constant, Ke.is calculated using molar concentrations. For gaseous reactions Ke another form of the equilibrium constant, Kp. is 71.6 at a temperature of 159 °C Calculate the value of Kp. calculated from partial pressures instead of concentrations. These two equilbrium constants Express your answer numerically are related by the equation where R 0.08206 L-atm/(K.mol), T is the absolute temperature, and An is the change in the number of of gas (sum moles products sum moles reactants). For example, consider the 2 (g) 2NH3(g) Hints My Answers Give Up Review Part Part B For the reaction K, 182x10-2 at a temperature of 31 C Calculate the value of Kr. Express your answer Submit MLAnswers Gave up Review Part Incorrect: Try Again; 5 attempts remaining erende Continue

Explanation / Answer

2H2S(g) 2H2(g) + S2(g)

initiall:   3.70×104 0 0

change: -2x +2x +x

equalibrium: 3.70×104 -2x 2x x

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

1.67×107 = (2x2 . x ) /  (3.70×104 - 2x)2

2x3 = 1.67×107 x( 3.70×104)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] = 3.70×104 - 2x

= 3.70×104 - ( 2 x 2.25 x 10-5)

   = 3.70×104 - 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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