268 SAMPLE EXAMINATION 2 o0 Question5 Consider the equilibriua reaction A mixtur

Question

268 SAMPLE EXAMINATION 2 o0 Question5 Consider the equilibriua reaction A mixture coataining 1.00 mole each of methane and water vapor is brought to equilibrium in a rigid 1.00 L flask at 800 K. The equilibrium mixture is found to contain 0.800 moles of Hy (a) What is the value of Kc for this reaction at 800 K? (b) If a 2.00 L flask is substituted for the original flask, what is the value of Kc when equilibrium is CH(g)+ 2 H,O(g) CO,(g)+4 H(g) flask at 800 K. The equilibrium mixture is found to contain 0 re-established at 800K?

Explanation / Answer

    CH4(g) + 2 H2O(g) ---------------- CO2(g) +   4 H2(g)

   1 mole       1 mole                         0              0

   -x                 -2x                             +x              + 4x

1.0-x            1.0-2x                           +x              +4x(0.800mole)

according to the given data

at equilibrium number of moles of H2 = 0.800 moles

i,e4x= 0.800

x= 0.2

at equlibrium number of moles of CH4 = 1.0-x= 1.0 - 0.2 = 0.8 moles

number of moles of H2O = 1.0-2x= 1.0-2x0.2= 0.6 moles

number of moles of CO2= x= .0.2 moles

volume = 1.0L

Concentration of CH4 = 0.8/1.0= 0.8M

concentration of H2O = 0.6/1.0= 0.6M

concentration of CO2 = 0.2/1.0 = 0.2M

concnetration of H2= 0.80/1.0= 0.8M

Kc = [CO2][H2]^4/[CH4][H2O]^2

Kc= (0.2)(0.8)^4/(0.8)(0.6)^2

Kc= 0.08192/0.288

Kc= 0.284

b) volume = 2.0L

[CH4] = 0.8/2.0=0.4M

[H2O]= 0.6/2.0= 0.3M

[CO2] = 0.2/2.0= 0.1M

]H2] = 0.8/2.0= 0.4M

Kc= (0.1)(0.4)^4/(0.4)(0.3)^2

Kc=0.00256/0.036 =0.071

Kc= 0.071

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