Question

When 0.514 g of biphenyl (c12h10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 oc to 29.4 oc. find δerxn for the combustion of biphenyl in kj/mol of biphenyl. the heat capacity of the bomb calorimeter, determined in a separate experiment is 5.86 kj/oc?

Answer 1

The solution for this problem is:Heat of combustion = heat capacity of calorimeter x ∆T
Q = 5.86kJ / ºC x 3.6ºC = 21.096kJ
0.514g / 154g/mole = 0.00334 moles
21.096kJ / 0.00334moles = 6.32x10^3kJ/mole is the heat capacity of the bomb that is determined in a separate experiment.

Answer 2

ΔErnx = -6384.55 kJ/mol

Step-by-step explanation:

A bomb calorimeter operates at a constant volume.

I.e ΔV = 0.

The first law of thermodynamics states that, energy can neither be created nor destroyed, but may be converted from one form to another.

Mathematically,

ΔU = ΔH + pΔV

Where ΔU = change in internal Energy, ΔH = Heat adsorbed or evolved and pΔV = work done by the system.

since a bomb calorimeter operates with constant volume,

∴ ΔV=0 as such, pΔV=0.

Then, ΔU = ΔH

This can be written as

ΔE = qv .

Where qv is the heat flowing out from the reaction towards the surrounding water at constant volume.

qv = 5.86 Kj/⁰c × (29.4-25.8)

qv = 5.86 × 3.6

qv = 21.096 kJ.

∴ ΔE of the solution = 21.096 kJ.

Because combustion is an exothermic reaction,

qv = -ΔE

and ΔErnx = -ΔE/ no of moles.

-ΔE = -21.096 kJ.

no moles of (C₁₂H₁₀) = mass of(C₁₂H₁₀)/molar mass of (C₁₂H₁₀)

mass of(C₁₂H₁₀) = 0.514g

molar mass of (C₁₂H₁₀) = (12×12) +(1×10) = 144+10 = 154g/mol.

∴ no of mole of (C₁₂H₁₀) = 0.514/154

≈ 0.0033 mol.

∴ΔErnx = -21.096/0.0033

ΔErnx ≈ -6386.55 kJ/mol

ΔErnx = -6384.55 kJ/mol