J.R. S. answered • 04/28/20

Ph.D. University Professor with 10+ years Tutoring Experience

1). To determine the heat capacity of the calorimeter (C_{cal}), you determine where the extra heat went that you didn't measure compared to what you should have measured theoretically.

Heat expected = 1.250 g x 24.30 kJ/g = 30.375 kJ

q = C_{cal}∆T

C_{cal} = q/∆T where ∆T is the change in the temperature = 47.60 - 25.00 = 22.60º

C_{cal }= 30.375 kJ/22.60º

C_{cal} = 1.344 kJ/degree

2). PCI_{5}(g) -> PCI_{3}(g) + Cl_{2}(g) ... TARGET EQUATION

Since PCl_{5} is on the left we reverse the second equation:

4PCl_{5} ==> P_{4} + 10Cl_{2} ∆Hº = +3438 kJ

Now just copy the first equation because it has PCl_{3} on the right side of the equation

P_{4} + 6Cl_{2} ==> 4PCl_{3} ∆Hº = -2439 kJ

Adding them together we have...

4PCl_{5 }+ P_{4} + 6Cl_{2} ==> P_{4} + 10Cl_{2} + 4PCl_{3} ∆Hº = 3438 kJ + -2439 kJ = 999 kJ

cancel P_{4} and Cl_{2} to end up with 4PCl_{5} ==> 4Cl_{2} + 4PCl_{3} ∆Hº = 999 kJ

Now divide by 4 to get the target equation...

PCl_{5} ==> Cl_{2} + PCl_{3} ∆Hº = 999 kJ/4 = 249.8 kJ