Question

a 6.55g sample of aniline (C6H5NH2 M.M=93.1 g/mol) was combusted in a bomb calorimeter with heat capacity of 14.25 kj/°C. if the initial temp. was 32.9°C use the information below to determine the value of final temperature of the calorimeter.                   4C6H5NH2 (l) + 35 O2 (g) -> 24 Co2 (g) + 14 H2o (g)+ 4NO2 (g) ◇H°RXN= -1.28 x 10^4 KJ

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

= 225.1 KJ

This heat is absorbed by calorimeter

use:

Q = C*(Tf-Ti)

225.1 KJ = 14.25 KJ/oC * (Tf - 32.9) oC

Tf - 32.9 = 15.796

Tf = 48.7 oC

Answer: 48.7 oC