Question

(a) Consider the reaction of hydrogen sulfide with methane, given below:

1 CH4(g) + 2 H2S(g) 1 CS2(g) + 4 H2(g) If CH4(g) is decreasing at the rate of 0.740 mol/s, what are the rates of change of H2S(g), CS2(g), and H2(g)? H2S(g)/t = mol/s CS2(g)/t = mol/s H2(g)/t = mol/s
(b) The decomposition reaction given below:
2 NH3(g) 1 N2(g) + 3 H2(g) is carried out in a closed reaction vessel. If the partial pressure of NH3(g) is decreasing at the rate of 327 torr/min, what is the rate of change of the total pressure in the vessel? Ptot /t = torr/min

Answer 1

a. H2S(g)/t = 1.48 mol/s

CS2(g)/t = 0.740mol/s

H2(g)/t = 2.96mol/s

b. Ptot /t = 981torr/min

Step-by-step explanation:

a. Based on the reaction:

CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)

1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2

If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s

CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s

The H2 is produced four times faster than CH4 is decreased, that is:

0.740mol/s * 4 = 2.96mol/s

b. With the reaction:

2 NH3(g) → N2(g) + 3 H2(g)

2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.

That means 2 moles of gas are consumed and 4 moles of gas are produced.

If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =

654torr/min

The rate of change of the total pressure is rate of reactants + rate of products:

654torr/min + 327torr/min =

981torr/min