Question

Calculate the enthalpy of formation of benzene at 25°C, if the enthalpy of
combustion of benzene, carbon and hydrogen at 25°C is -3276.0, -394.8 and
-286.86 kJ respectively at 25°C.​

Answer 1

Answer:

49.4

Explanation:

Given Thermochemical Equations:

(i) C6H6 + (15/2) O2 ----> 6CO2 + 3H2O ΔH = -3266.0 kJ

(ii) C + O2 ----> CO2 ΔH = -393.1 kJ

(iii) H2 + (1/2) O2 ---> H2O ΔH = -286.0 kJ

Now, we want enthalpy of formation of benzene, so we require C6H6 on the product side.

Reverse the equation (i); multiply the equation (ii) by 6 in order to get 6 CO2 ; multiply the equation (iii) by 3 to get 3 H2O. So, we will have:

(iv) 6CO2 + 3H2O ----> C6H6 + (15/2) O2 ΔH = + 3266.0 kJ

(v) 6C + 6O2 ----> 6CO2 ΔH = -2358.6 kJ

(vi) 3H2 + (3/2) O2 ---> 3H2O ΔH = -858.0 kJ

Notice that on reversing the equation (i) the sign of ΔH changes as well.

Now, observe that the 6CO2 in reaction (iv) on the reactant side cancels out with 6CO2 in reaction (v) in the product side.

Similarly, H2O and O2 also cancels out, as they are present in same molar quantities on both sides. So, we are left with:

6C + 3H2 ----> C6H6

and on adding their respective ΔH values we get the enthalpy of formation of benene = + 3266.0 kJ + (-2358.6 kJ) + (-858.0 kJ) = 49.4

Answer 2

The enthalpy of formation of benzene at 25°C is 46.62 kJ.

Given,

The enthalpy of combustion of benzene, carbon and hydrogen at 25°C is -3276.0, -394.8 and -286.86 kJ respectively at 25°C.​

To find,

Calculate the enthalpy of formation of benzene at 25°C

Solution,

We know that,

$C_{6}H_{6} + \frac{15}{2}O_{2}$   --------> $6CO_{2} + 3H_{2}O$

Δ$H_{reaction}$ = ∑$Hf_{product}$ - ∑$Hf_{reactant}$

      -3276 =  [ 6(-394.8) + 3(-286.86) ] - [$Hf_{C6H6}$ + $\frac{15}{2}$ x 0]

      -3276 = -2368.8 - 860.58 - $Hf_{C6H6}$

    $Hf_{C6H6}$ = -3229.38 + 3276 = 46.62

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