When .514 g of C12H10 undergoes combustion in a bomb calorimeter, the temperature increases from 25.8oC...

Question

Question

When .514 g of C₁₂H₁₀ undergoes combustion in a bomb calorimeter, the temperature increases from 25.8^oC to 29.4^oC. The heat capacity of the bomb and all of its contents is 5.86 kJ/^oC.

Balanced reaction: 2 C₁₂H₁₀ (s) + 29 O₂ (g) → 24 CO₂ (g) + 10 H₂O (g)

If ΔE and ΔH are not equal, which one is more negative? Why?
* Please type the answers, don't write them down on a paper

Answer 1

Answer #1

A.

q_v = 5.86 KJ/⁰C _* 3.6 ^₀C _* 154 g/mol / 0.514 g

q_v= - 6320.591 KJ/mol (heat of combustion always have a negative sign )

b. has been calculated as in combuustion bond dissociates so bond energy i.e., enthalpy of formation of bond is calculated.

c. because in the experiment = 0.

and

Answer 2

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