Help! Please! Heat Capacity of Calorimeter & Hess's Law

Question

1. You used a bomb calorimeter, or constant-volume calorimeter, but misplaced the information of heat capacity of the calorimeter. If you burn 1.250 g of an organic compound, with, ΔE = -24.30kJ/g, and measured a temperature change from 25.00^oC to 47.60^oC, what must the heat capacity of the calorimeter be?

THINK I GOT 2 FEEL FREE TO DO IT SO I CAN CHECK BUT I JUST NEED !

2.Using the Hess's Law and the manipulations of the following thermochemical equations:

P₄(s) + 6Cl₂(g) -> 4PCl₃(g) ΔH^o : -2439 kJ

P₄(s) + 10Cl₂(g) -> 4PCl₅(g) ΔH^o : -3438 kJ

Determine the enthalpy oI reaction for the following reaction:

PCI₅(g) -> PCI₃(g) + Cl₂(g) ΔH^o = ?

Please make answers as detailed as possible. I thought I knew what I was doing but I'm so lost. Please help I have to turn this is in soon and have been working on it forever.

Answer 1

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). PCI₅(g) -> PCI₃(g) + Cl₂(g) ... TARGET EQUATION

Since PCl₅ is on the left we reverse the second equation:

4PCl₅ ==> P₄ + 10Cl₂ ∆Hº = +3438 kJ

Now just copy the first equation because it has PCl₃ on the right side of the equation

P₄ + 6Cl₂ ==> 4PCl₃ ∆Hº = -2439 kJ

Adding them together we have...

4PCl₅ + P₄ + 6Cl₂ ==> P₄ + 10Cl₂ + 4PCl₃ ∆Hº = 3438 kJ + -2439 kJ = 999 kJ

cancel P₄ and Cl₂ to end up with 4PCl₅ ==> 4Cl₂ + 4PCl₃ ∆Hº = 999 kJ

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

PCl₅ ==> Cl₂ + PCl₃ ∆Hº = 999 kJ/4 = 249.8 kJ