(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

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.