Calculating Heat of Combustion

[ben4life86]

A 0.196 g sample of quinone (C6H4O2) is burned in a bomb calorimeter that has a heat capacity of 1.56 kJ/C. What is the heat of combustion (kJ/mol) if the temperature increases by 3.2 C?

-I have no idea how to do this problem...

 

[Gokul43201]

Look at your text. It has examples of similar problems.

 

[ravenprp]

Calculating the heat of combustion involves using the formula Q = mCΔT, where Q is the heat released, m is the mass of the substance, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

First, we need to convert the mass of quinone from grams to moles. Since the molar mass of quinone is 108.1 g/mol, the moles of quinone in 0.196 g is 0.00181 mol.

Next, we can plug in the values into the formula Q = mCΔT. The mass of quinone is 0.196 g, the heat capacity of the calorimeter is 1.56 kJ/C, and the change in temperature is 3.2 C. This gives us:

Q = (0.196 g)(1.56 kJ/C)(3.2 C) = 0.98 kJ

We now have the heat released, but we need to find the heat of combustion (kJ/mol). To do this, we can use the mole ratio between quinone and the heat released. Since 0.00181 mol of quinone released 0.98 kJ, we can set up a proportion:

0.00181 mol / 0.98 kJ = 1 mol / x kJ

Solving for x, we get x = 540.88 kJ/mol. Therefore, the heat of combustion for quinone is 540.88 kJ/mol.