# Please answer immediately

We have to calculate the enthalpy of formation of ethyl alcohol, so we have to calculate ΔH for the following reaction:

2C + 6H + 1/2 O_{2} -----------------------> C_{2}H_{5}OH ..........................................................(a)

Combustion of ethyl alcohol is given by:

C_{2}H_{5}OH + 3O_{2} ---------> 2CO_{2} + 3H_{2}O ΔH = 1380.7 KJ/mol ....................................(i)

Formation of CO_{2} is given by:

C + O_{2} ---------> CO_{2} ΔH = 394.5 KJ/mol ..........................................................(ii)

Formation of H_{2}O is given by:

H_{2} + 1/2 O_{2} ---------> H_{2}O ΔH = 286.6 KJ/mol .....................................................(iii)

To Obtain equation (a), we have to multiply equation (ii) by 2 and equation (iii) by 3 and then add these two equations, we get

2C + 3H_{2} + 7/2 O_{2} -------------> 2CO_{2} + 3H_{2}O ΔH = 1648.8 KJ/mol ..............................(iv)

Now subtract equation (i) from equation (iv), we will get

2C + 3H_{2} + 1/2 O_{2} -------------> C_{2}H_{5}OH ΔH = 268.1KJ/mol

So, enthalpy of formation of ethyl alcohol is 268.1KJ/mol.

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