Use the standard reaction enthalpies given below to determine δh°rxn for the following reaction:

p4(g) + 10 cl2(g) → 4pcl5(s) δh°rxn = ?
given:
pcl5(s) → pcl3(g) + cl2(g) δh°rxn = +157 kj
p4(g) + 6 cl2(g) → 4 pcl3(g) δh°rxn = -1207 kj

(a) -1050. kj
(b) -2100. kj
(c) -1364 kj
(d) -1786 kj
(e) -1835 kj

(E) -1835 kJ

Explanation:

Use Hess law to answer this question by manipulating algebraically the equations :

Eq .1   PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157

Eq.2    P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

to arrive to the equation in our problem:

P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?

Notice PCl₅ is the product in our equation and in Eq. 1 is a reactant so we need to reverse it. We need also to multiply it by 4 since that is the number required in the question.

P₄ in equation 2 is a reactant and is also a reactant in our desired equation also it is 1 mol in both. That is telling us to leave it as it is and we can then proceed to add it  the the  -4 times Eq1 and check everything is right when we add them together.

4PCl3(g) + 4Cl2(g)  → 4PCl5(s)   ΔH°rxn = -4 x (157 kJ)  (change the sign)

P4(g) + 6 Cl2(g) → 4 PCl3(g)       ΔH°rxn = -1207 kJ

P4(g) + 10 Cl2(g) → 4PCl5(s)        ΔH°rxn =   -4 x (157 kJ) +   -1207 kJ

                                                     ΔH°rxn =   -1835 kJ

See how the PCl3 cancel and the Cl add correctly.

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.

girl imma just put my comment here because i need a couple point okurrttt