Question

A mixture of 3.00 moles of methane and 2.00 mol of divisor sulfide (hydrogen sulphide) initially reacted at 1000 K according to the following reaction: CH4 (g) + 2 H2S (g) ↽ - ⇀ CS2 (g) + 4 H2 (g) In equilibrium, the hydrogen partial pressure was 20.0 kPa and the total pressure 92.0 kPa.

a) What is the total molar amount of reaction in balance? (indicate to four decimal places)

b) Calculate the partial pressures of the various substances in the reaction. Report the replies to two decimal places, in kPa
c) Calculate the value of equilibrium constant Kp. Report the answer to two decimal places with the unit (kPa)^2
d) What is the total volume of the gas mixture? Impose the answer in m3 to the 3rd decimal place.

Answer 1

The Given Reaction is

CH4 (g) + 2 H2S (g) ↽ - ⇀ CS2 (g) + 4 H2 (g)

Moles at time t=0 3 2 0 0 total moles= 5

Moles at time t=t 3-x 2-2x x 4x total moles= 5+2x

mole fraction of hydrogen = 4x / (5+2x)

partial pressure = mole fraction * total pressure

20 = 4x / (5+2x) * 92

20* (5+2x) = 4x * 92

100 + 40x = 368x

100 = 328x

0.3049 = x

With the above information, we can find answers to all the parts

Part (a)

total moles = 5+2x

= 5.6097

Part (b)

To solve this we can use the formula partial pressure = mole fraction * total pressure

Part (c)

by using the above formula the answer can be obtained

Part (d)

We can use the ideal gas equation PV = nRT to solve this part

total pressure P = 92 kPa

total moles n = 5.6097

Gas Constant R = 8.314 J/(mol.K)

Temperature T = 1000K

Hence we get V = 0.507 m3