Question

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. All work must be shown.
4 C6H5NH2(l) + 35 O2(g) → 24 CO2(g) + 14 H2O(g) + 4 NO2(g) ΔH°rxn = -1.28 × 10^4 kJ

A) 257°C B) 48.7°C C) 41.9°C D) 46.6°C E) 931°C

Answer 1

Molar mass of C6H5NH2 = 6*MM(C) + 7*MM(H) + 1*MM(N)

= 6*12.01 + 7*1.008 + 1*14.01

= 93.126 g/mol

mass of C6H5NH2 = 6.55 g

we have below equation to be used:

number of mol of C6H5NH2,

n = mass of C6H5NH2/molar mass of C6H5NH2

=(6.55 g)/(93.126 g/mol)

= 7.033*10^-2 mol

Since delta H is negative, heat is released

when 4 mol of C6H5NH2 reacts, heat released = 12800.0 KJ

So,

for 7.033*10^-2 mol of C6H5NH2, heat released = 7.033*10^-2*12800.0/4 KJ

= 2.251*10^2 KJ

this heat is absorbed by calorimeter

use:

Q = Ccal * (Tf-Ti)

225.1 = 14.25*(Tf-32.9)

Tf-32.9 = 15.8

Tf = 48.7 oC

Answer: 48.7 oC