(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

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

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

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun