A 0.56 g sample of liquid C6H12 was combusted completely using excess oxygen inside a bomb (constant volume) calorimeter, with the products being carbon dioxide and liquid water. The calorimeter's heat capacity is 5.36 kJ °C-1. If the temperature inside the calorimeter increased from 25.0 °C to 29.2 °C, determine ΔrH for this reaction in kJ mol-1 (with respect to C6H12) at 298 K. Do not worry about how realistic the final answer is. You have 5 attempts at this question. TIP: To report an answer in scientific notation, enter it using the format "2.3E4", which means "2.3 x 104" (without the quotation marks)

-3.4 × 10³ kJ/mol

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

where,

: heat absorbed by the calorimeter

: heat capacity of the calorimeter

: change in the temperature

Step 2: Calculate the heat released by the combustion

According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the combustion is zero.

Step 3: Calculate the standard enthalpy of the combustion

22.5 kJ are released when 0.56 g of C₆H₁₂ (MW 84.16) is combusted. The standard enthalpy of the combustion (ΔH°c) is:

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.