Question

The combustion of one mole of benzene takes place at 298 k and 1 atm. After combustion, co2(g) and h2o (1) are produced and 3267.0 kj of heat is liberated. Calculate the standard enthalpy of formation, f h0 of benzene. Standard enthalpies of formation of co2(g) and 2h o(l)are 393.5 kj mol1 and 285.83 kj mol1 respectively.

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student-name Anshija asked in Chemistry

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ΔfHΘof benzene. Standard enthalpies of formation of CO2(g) and H2O(l) are -393.5 kJ mol1and 285.83kJ mol1respectively.

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student-name Anita Dahiya answered this

2488 helpful votes in Chemistry, Class XII-Science

The formation reaction of benezene is given by :

6C (graphite) + 3H2 g → C6H6 (l) ;

ΔfHΘ = ? … —————————————–(i)

The enthalpy of combustion of 1 mol of benzene is :

C6H6 (l) + 15/2 O2 → 6CO2 (g) + 3H2O (l);

ΔcHΘ = -3267kJ mol−1 … ——————————–(ii)

The enthalpy of formation of 1 mol of CO2(g) :

C (graphite) + O2(g) → CO2 g ;

ΔfHΘ = -393.5kJ mol−1 … ———————————————(iii)

The enthalpy of formation of 1 mol of H2O(l) is :

H2 (g) + 1/2 O2 (g) → H2O (l);

ΔfHΘ = -285.83kJ mol−1 … ——————————————–(iv)

multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get:

6C (graphite) + 6O2 g → 6CO2 g ;

ΔfHΘ = -2361kJ mol−1

3H2 (g) + 3/2 O2 (g) → 3H2O (l);

ΔfHΘ = -857.49kJ mol−1

Summing up the above two equations :

6C (graphite) + 3H2 (g) + 15/2 O2 g → 6CO2 (g) + 3H2O(l);

ΔfHΘ = −3218.49 kJ mol-1 … ————————–(v )

Reversing equation (ii);

6CO2 (g) + 3H2O(l) → C6H6(l) + 15/2 O2;

ΔfHΘ = −3267.0 kJ mol-1 … ————————–(vi )

Adding equations (v) and (vi), we get

6C (graphite) + 3H2 (g) → C6H6(l);

ΔfHΘ = 48.51 kJ mol-1

Answer 2

Answer: The enthalpy of the formation of $C_6H_6$ is coming out to be 1584.99 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(12\times \Delta H^o_f_{(CO_2)})+(6\times \Delta H^o_f_{(H_2O)})]-[(2\times \Delta H^o_f_{(C_6H_6)})+(15\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(H_2O)}=285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2)}=393.5kJ/mol\\\Delta H^o_{rxn}=3267.0kJ$

Putting values in above equation, we get:

$3267.0=[(12\times (393.5))+(6\times (285.83))]-[(2\times \Delta H^o_f_{(C_6H_6)})+(15\times (0))]\\\\\Delta H^o_f_{(C_6H_6)}=1584.99kJ/mol$

Hence, the enthalpy of the formation of $C_6H_6$ is coming out to be 1584.99 kJ/mol.