# 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.

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

According to the reaction equation:

4C6H5NH2 +35 O2  → 24CO2(g) + 14 H2O(g) +  4NO2(g)  ΔH = -1.28x10^4 KJ

from this equation, we can need 4 mol of aniline to give 1.28 x 10^4 KJ ΔH

first, we need to get moles of aniline = mass/molar mass

= 6.55 g / 93.13g/mol

=0.07 mol

the heat generated = moles of aniline / 4 mol * ΔH

= 0.07mol / 4 mol * 1.28 x 10^4KJ

= 224 KJ
when ΔT = Q/C

when we have C the heat capacity = 14.25 KJ/°C

ΔT = 224 / 14.25

= 15.7°C

∴ Tf = Ti + ΔT

= 32.9 °C + 15.7°C

=  48.6 °C

Related Posts