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Question: Heat of combustion of benzoic acid \[({{C}_{6}}H_{5}^{{}}COOH)\] at constant volume at \(25{}^\circ ...

Heat of combustion of benzoic acid (C6H5COOH)({{C}_{6}}H_{5}^{{}}COOH) at constant volume at 2525{}^\circ is 3233KJ/mole3233KJ/mole. When 0.5g0.5g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by 0.53C.0.53{}^\circ C. Now in the same bomb calorimeter 1g1g of C2H6{{C}_{2}}{{H}_{6}} burnt then temperature increased by 2.04C.2.04{}^\circ C. ΔH\Delta Hfor combustion of C2H6{{C}_{2}}{{H}_{6}} is?

Explanation

Solution

We know that Heat of formation is the heat change (either evolved or absorbed) accompanying the formation of one mole of a substance from its elements under a given condition of temperature and pressure. Also, the heat of combustion is the amount of heat liberated when one mole of a substance is completely burnt or oxidized in air

Complete step by step solution:
We are given the heat of combustion of carbon, hydrogen, and acetylene and asked to find the Heat of combustion of benzoic acid (C6H5COOH)({{C}_{6}}H_{5}^{{}}COOH) . So since we are given the heat of formation, let’s find out what that really means. The heat of combustion is the heat liberated when it is completely burnt or oxidized in air. So let’s make carbon to oxidise in air.
ΔHcamb\underset{camb}{\mathop{\Delta H}}\,​ for (C6H5COOH)3233KJ/mole({{C}_{6}}H_{5}^{{}}COOH)-3233KJ/mole
moles in OO System =0.5122=1244 mol=\dfrac{0.5}{122}=\dfrac{1}{244}~mol
Therefore, heat ecvaluted =(1244×3233)kJ=\left( \dfrac{1}{244}\times 3233 \right)kJ and it get specific heat for bomb calanmeter be
5×0.53=3233244kJ\therefore 5\times 0.53{}^\circ =\dfrac{3233}{244}kJ
8=3233244×0.53=25kJ/ C\Rightarrow 8=\dfrac{3233}{244\times 0.53}=25kJ/~{}^\circ C
Therefore, when temp included by 2.04 C2.04~{}^\circ C
Thus, Heat supplied =2.04×25=51kJ=2.04\times 25=51kJ
Therefore, 54 kJ 54~kJ~ is released when 1g C2H6 1g\text{ }{{C}_{2}}{{H}_{6}}~is burnt; i.e. 130mol\dfrac{1}{30}mol is;
Therefore, for 11 mol, heat is 51×30=1530 kJ/mol.51\times 30=1530~kJ/mol.

Note:
Remember that the efficiency of fuel means the output the fuel gives compared to the input energy we have to give to burn the fuel. A fuel that is highly efficient in nature is said to have a high calorific value is the amount of heat that is released by a fuel, on complete combustion, when a unit weight, rather than unit volume of it is used up.