ThermoChemistry

Question

Liquid hydrogen peroxide is an oxidizing agent in rocket fuel mixtures because it decomposes to form oxygen, which is then used in the combustion of the rocket fuel. The reaction is:

2 H₂O₂ (l) → 2 H₂O(l) + O₂(g)    ΔH = -196.1 ^kJ/_mol

If a 36.99 g sample of hydrogen peroxide is placed in a bomb calorimeter containing 320.5 g of water at 21.6 ˚C and allowed to react, what will be the final temperature of the water? Assume that the calorimeter absorbs a negligible amount of heat.

 

Answer 1

Use the formula q = mC∆T

q = heat (see below for calculation of heat)

m = mass of water = 320.5 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = ?

To find the heat generated by the reaction, we use the fact that 196.1 kJ / mol is generated by 2 moles of 2H2O2 as shown in the balanced equation. So, first we find the number of moles of H2O2 that we have.

moles H2O2 = 36.99 g x 1 mol / 34.01 g = 1.0876 moles

Energy = heat = q = 1.0876 mols H2O2 x 196.1 kJ/mol = 213.3 kJ = 213300 J

Now substitute into the equation for q:

q = 213300 J = mC∆T = (320.5 g)(4.185 J/gº)(∆T)

∆T = 213300 J / (320.5 g)(4.185 J/gº)

∆T = 159.1º

Since the sign of ∆H is negative, this means heat is generated by the reaction so the temperature will increase.

Final temperature = 21.6º + 159.1º = 180.7ºC