Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

    4. Answer the following questions.

    ix. Calculate standard enthalpy of reaction, Fe2O3(s) + 3CO(g) 2 Fe(s) + 3CO2(g), from the following data. ΔfH0(Fe2O3) = -824 kJ/mol, ΔfH0(CO) = -110 kJ/mol, ΔfH0(CO2) = -393 kJ/mol
    Ans. : (-25 kJ)

    Answer:-

    Given:

    • \(\Delta_f H^\circ (\text{Fe}_2\text{O}_3) = -824 \, \text{kJ/mol}\)
    • \(\Delta_f H^\circ (\text{CO}) = -110 \, \text{kJ/mol}\)
    • \(\Delta_f H^\circ (\text{CO}_2) = -393 \, \text{kJ/mol}\)

    To find: Standard enthalpy of the given reaction (\(\Delta_r H^\circ\))

    Formula: \(\Delta H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})\)

    Calculation:

    The reaction is

    \[ \text{Fe}_2\text{O}_3(\text{s}) + 3\text{CO}(\text{g}) \rightarrow 2\text{Fe}(\text{s}) + 3\text{CO}_2(\text{g}) \]

    \(\Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})\)

    \(= [2\Delta_f H^\circ (\text{Fe}) + 3\Delta_f H^\circ (\text{CO}_2)] - [\Delta_f H^\circ (\text{Fe}_2\text{O}_3) + 3\Delta_f H^\circ (\text{CO})]\)

    \(= [0 + 3 \, \text{mol} \times (-393 \, \text{kJ/mol})] - [1 \, \text{mol} \times (-824 \, \text{kJ/mol}) + 3 \, \text{mol} \times (-110 \, \text{kJ/mol})]\)

    \(= -1179 + 824 + 330\)

    \(= -25 \, \text{kJ}\)

    The standard enthalpy of the given reaction is \(-25 \, \text{kJ}\).

    Q 4.8
    Q 4.9
    Q 4.10

    Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution