A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon.

Answer : The enthalpy change per mole of hydrocarbon is, 269 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

= heat absorbed by the water

= specific heat of calorimeter =

= specific heat of water =

= mass of water = 925 g

= change in temperature =

Now put all the given values in the above formula, we get:

Now we have to calculate the enthalpy change per mole of hydrocarbon.

where,

= enthalpy change = ?

q = heat released = 134.5 kJ

n = moles of hydrocarbon = 0.500 mol

Therefore, the enthalpy change per mole of hydrocarbon is, 269 kJ/mole

[1] P4 + [3] O2=[2] P2O3

Explanation:

you can first solve for amount of O2 in the equation because reactant had O2 and product had O3, so the amount of O2 should be 3 and amount of O3 should be 2, which gets u to 6 moles of oxygen for both reactant and product.  then you can balance the P4, and 1 mols of P4 = 2 moles of P2, which matches with the amount of mols for O3 as well. therefore you can get P4 + 3O2 = 2P2O3