Question

11.9 g sample of aniline (C6H5NH2, molar mass = 93 g/mol) was combusted in a calorimeter with sufficient of oxygen. If the calorimeter heat capacity is 10.0 kJ/K, how many Kelvin does the temperature increase due to this sample combustion?

4 C6H5NH2 (l) + 35 O2 (g) --> 24 CO2 (g) + 14 H2O (g) + 4 NO2 (g)

ΔHrxn=−12800kJ

Answer 1

The temperature increases in 40.9K

Step-by-step explanation:

Based on the reaction, 4 moles of aniline in combustion releasing 12800kJ to the surroundings, in this case, to the calorimeter.

To solve this question we must find the heat that releases the sample of aniline finding the moles in the sample.

With the heat released we can find the increase in Kelvin due the combustion as follows:

Moles aniline:

11.9g * (1mol / 93g) = 0.128 moles aniline

Heat released:

0.128 moles Aniline * (12800kJ / 4 moles Aniline) = 409kJ are released

Increasing in temperature:

For each 10.0kJ, the calorimeter increases its temperature in 1K. For 409kJ:

409kJ * (1K / 10.0kJ) =

The temperature increases in 40.9K