A6.55 g sample of aniline (c6h5nh2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25 kj/°c. if the initial temperature was 32.9°c, use the information below to determine the value of the final temperature of the calorimeter.
4 c6h5nh2(l) + 35 o2(g) → 24 co2(g) + 14 h2o(g) + 4 no2(g) δh°rxn= -1.28 x 104 kj

Final temperature = 48.6867 °C

Explanation:

The expression for the heat of combustion in bomb calorimeter is:

ΔH = -C ΔT

where,

ΔH is the enthalpy of the reaction

C is the heat capacity of the bomb calorimeter

ΔT is the temperature change

Given in the question:

ΔH°rxn = -1.28 x 10⁴ kJ

From the balanced reaction, 4 moles of aniline reacting with oxygen. Thus, enthalpy change of the reaction in kJ/mol is:

ΔH = -12800 kJ/4 = -3200kJ/mole

Given:

Mass of aniline combusted = 6.55 g

Molar mass of aniline = 93.13 g/mol

Thus moles of aniline = 6.55 / 93.13 moles = 0.0703 moles

The total heat released from 0.0703 moles of aniline is

ΔH = -3200kJ/mole x 0.0703 moles = -224.96 kJ

Given: Heat capacity of calorimeter is 14.25 kJ/°C

T₁ (initial) = 32.9°C

T₂ (final) = ?

From the above formula:

-224.96 kJ = -14.25kJ/°C (T₂ - 32.9)

Solving for T₂ , we get:

T₂ = 48.6867 °C

it is stable as it is in group 0 which have full outer shells of electrons

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