A 6.55 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)

Question

Clara Rice

Chemistry

11 April, 22:10

A 6.55 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)

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Valentin · Answer 1

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