Question

When 10gm of $\text{Al}$ is used for reduction in each of the following alumino thermic reactions, which reaction would generate more heat and by how much?
(a) $\text{2Al}\,\text{+}\,\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\text{+}\,\text{2Cr}$
(b) $\text{2Al}\,\text{+}\,F{{e}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\text{+}\,\text{2Fe}$
Standard heat of formation of $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$,$\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ are $\text{-1676kJ}$,$\text{-1141kJ}$ and $\text{-822}\text{.2kJ}$, respectively.
(A) Reaction (b) produces more heat by $59.04\text{kJ}$
(B) Reaction (b) produces less heat by $59.04\text{kJ}$
(C) Reaction (a) and (b) produces equal heat
(D) None of these

Answer 1

Standard heat of formation is the enthalpy change (evolved or absorbed) when one mole of a substance is formed from it element in their most abundant naturally occurring form or in their standard and stable form.

Complete Solution :
In these questions firstly we calculate the generated heat by individual reaction and then calculate the difference of heat energies.
According to the question standard heat of formation of $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $\text{-1676kJ}$, and standard heat of formation of $\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $\text{-1141kJ}$.
So, $\begin{aligned} & \text{2Al}\,\text{+}\,\dfrac{3}{2}{{\text{O}}_{\text{2}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\,\,;\,\,\,\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}=\,-16\text{76kJ}\,\, \\\ & \text{54gm}\,\,\text{48gm}\,\,\,\,\,\,\,\text{102gm} \\\ \end{aligned}$

- From unitary method if

& \text{54gm}\,\,\text{Al}\,\,\text{will}\,\,\text{produces}\,\,\,\,=\,\,\,\,-16\text{76kJ}\,\text{energy} \\\ & \text{so,}\,10\text{gm}\,\,\,\text{Al}\,\,\,\text{will}\,\text{produces}\,\,\,=\,\,\,\dfrac{-167\text{6kJ}}{54}\,\times \,\,10\text{gm}\, \\\ & \,=\,\,\,-310.\text{4kJ} \end{aligned}$$ So we get for 10gm $\text{2Al}\,\text{+}\,\dfrac{3}{2}{{\text{O}}_{\text{2}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\,\,;\,\,\,\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}}=\,-310\text{kJ}\,...........(i)$ $\text{2Cr}\,\text{+}\,\dfrac{3}{2}{{\text{O}}_{\text{2}}}\,\to \,\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\,\,\,\text{;}\,\,\,\Delta {{\text{H}}_{f}}=\,-11\text{41kJ}\,\,.........\text{(ii)}$ From reversing equation (ii) and addition of equation (i) and equation (ii), we get $$\begin{aligned} & \text{2Al}\,\text{+}\,\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\text{+}\,\text{2Cr}\,;\, \\\ & \Delta {{H}_{f}}=\,1141\text{kJ}-310.4\text{kJ} \\\ & \Delta {{H}_{f}}=\,+830.6\text{kJ} \end{aligned}$$ In the same manner - According to the question standard heat of formation of $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $\text{-1676kJ}$, and standard heat of formation of $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $\text{-822}\text{.2kJ}$. So we get for 10gm of aluminium $\text{2Al}\,\text{+}\,\dfrac{3}{2}{{\text{O}}_{\text{2}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\,\,;\,\,\,\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{f}}} = \,-310\text{kJ}\,...........(iii)$ Heat of formation for $\text{(F}{{\text{e}}_{2}}{{\text{O}}_{3}}\text{)}$ $\text{2Fe}\,\text{+}\,\dfrac{3}{2}{{\text{O}}_{\text{2}}}\,\to \,\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\,\,\,\text{;}\,\,\,\Delta {{\text{H}}_{f}}=\,-8.22.2\text{kJ}\,\,.........\text{(iv)}$ From reversing equation (iv) and addition of equation (i) and equation (iv), we get $$\begin{aligned} & \text{2Al}\,\text{+}\,F{{e}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\to \,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\text{+}\,\text{2Fe}\,; \\\ & \,\Delta {{H}_{f}}=\,822.2\text{kJ}-310.4\text{kJ} \\\ & \Delta {{H}_{f}}=\,+511.8\text{kJ} \end{aligned}$$ So difference of absorbed energies in both (a) and (b) reaction – $$\begin{aligned} & \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{absorb}}}\,\,\,=\,\,\Delta {{H}_{a}}-\,\Delta {{H}_{b}} \\\ & \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{absorb}}}\,\,\,=\,+830.6\text{kJ}\,-\,511.8\text{kJ} \\\ & \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{absorb}}}=\,\,+318\text{kJ} \end{aligned}$$ Since, reaction (a) will absorb $$+318\text{kJ}$$ more energy than reaction (b). **So, the correct answer is “Option D”.** **Note:** The negative sign indicates that reaction is exothermic or heat is given out during reaction and therefore, temperature of the solution increases. While a positive sign indicates that reaction is endothermic and therefore, temperature of the solution decreases.