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Jakob Littel

Feb 20, 2025

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

2 Answers

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Hermina Ferry

Feb 20, 2025

Final temperature = 48.6867 °CExplanation:The expression for the heat of combustion in bomb calorimeter is:ΔH = -C ΔT where,ΔH is the enthalpy of the reactionC is the heat capacity of the bomb calorimeter ΔT is the temperature changeGiven 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/moleGiven:Mass of aniline combusted = 6.55 gMolar mass of aniline = 93.13 g/molThus moles of aniline = 6.55 / 93.13 moles = 0.0703 molesThe total heat released from 0.0703 moles of aniline is ΔH = -3200kJ/mole x 0.0703 moles = -224.96 kJGiven: Heat capacity of calorimeter is 14.25 kJ/°CT₁ (initial) = 32.9°CT₂ (final) = ?From the above formula:-224.96 kJ = -14.25kJ/°C (T₂ - 32.9)Solving for T₂ , we get:T₂ = 48.6867 °C... Show More
H
Hermina Ferry

Feb 20, 2025

According to the reaction equation:4C6H5NH2 +35 O2  → 24CO2(g) + 14 H2O(g) +  4NO2(g)  ΔH = -1.28x10^4 KJfrom this equation, we can need 4 mol of aniline to give 1.28 x 10^4 KJ ΔHfirst, we need to get moles of aniline = mass/molar mass                                                              = 6.55 g / 93.13g/mol                                                              =0.07 molthe heat generated = moles of aniline / 4 mol * ΔH                                 = 0.07mol / 4 mol * 1.28 x 10^4KJ                                = 224 KJwhen ΔT = Q/C when we have C the heat capacity = 14.25 KJ/°CΔT = 224 / 14.25     = 15.7°C∴ Tf = Ti + ΔT       = 32.9 °C + 15.7°C      =  48.6 °C... Show More

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