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)
ΔH°_rxn= -1.28 × 10⁴ kJ

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat produced in the combustion reaction. We can use the given ΔH°rxn value and the stoichiometry of the balanced equation to calculate the heat produced per gram of aniline combusted:

ΔHcomb = ΔH°rxn/n = (-1.28 × 10^4 J/mol) / (93.13 g/mol) = -137.3 J/g

Next, we can use the heat capacity of the calorimeter to calculate the temperature change caused by the heat produced in the combustion reaction:

q = CΔT

where q is the amount of heat produced by the reaction, C is the heat capacity of the calorimeter, and ΔT is the temperature change.

Rearranging this equation, we get:

ΔT = q / C = (-137.3 J/g) / (14.25 J/°C) = -9.64°C

This means that the final temperature of the calorimeter is:

Tf = Ti + ΔT = 32.9°C - 9.64°C = 23.26°C

However, we need to convert this temperature from Celsius to Kelvin to match the answer choices:

Tf (K) = Tf (°C) + 273.15 K = 23.26°C + 273.15 K = 296.41 K

Round to one decimal place:

Tf = 48.7°C.

Therefore, the correct answer is C.

A 655 G Sample of Aniline (C6H5NH2,molar Mass = 93

A 655 G Sample of Aniline (C₆H₅NH₂,molar Mass = 93