The Δf H of AO (g), AO2(g), B2O(g) and B2O4(g) is respectively -110, -393, +81 and +9.7 kJ mol-1. What is Δr H (in kJ mol-1) of the following reaction at 298 K? B2O4(g) + 3AO(g) → B2O(g) + 3AO2(g)

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q128 – AP-EAMCET 2026

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Home Chemistry Chemical ThermodynamicsQuestion #291

The

Δ_{f} H^{Θ}

of AO (g), AO

_{2}

(g), B

_{2}

O(g) and B

_{2}

_{4}

(g) is respectively

- 110

- 393

+ 81

and

+ 9.7 {kJ mol}^{- 1}

. What is

Δ_{r} H^{Θ}

(in

{kJ mol}^{- 1}

) of the following reaction at 298 K?\B

_{2}

_{4}

(g) + 3AO(g)

\to

_{2}

O(g) + 3AO

_{2}

(g)

Clue 1 of 3: Enthalpy of Reaction from Enthalpies of Formation

The standard enthalpy change of a reaction, $Δ_{r} H^{Θ}$ , can be calculated using the standard enthalpies of formation, $Δ_{f} H^{Θ}$ , of the reactants and products.

🧠 Full Solution Path

Step 1: Identify Enthalpies of Formation and Reaction✦ Active

The standard enthalpies of formation ( $Δ_{f} H^{Θ}$ ) for the participating compounds are given:

Δ_{f} H^{Θ} (AO(g)) = - 110 {kJ mol}^{- 1}

Δ_{f} H^{Θ} ({AO}_{2} (g)) = - 393 {kJ mol}^{- 1}

Δ_{f} H^{Θ} (B_{2} O(g)) = + 81 {kJ mol}^{- 1}

Δ_{f} H^{Θ} (B_{2} O_{4} (g)) = + 9.7 {kJ mol}^{- 1}

The reaction is:

B_{2} O_{4} (g) + 3 AO(g) \to B_{2} O(g) + 3 {AO}_{2} (g)

We need to calculate the standard enthalpy change of the reaction, $Δ_{r} H^{Θ}$ .

💡 Teacher's Secret Hint

Ensure you correctly identify which compounds are reactants and which are products, and their respective stoichiometric coefficients from the balanced equation.

Step 2: Apply Hess's Law○ Expand

Hess's Law states that the standard enthalpy change of a reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants, each multiplied by their stoichiometric coefficients:

Δ_{r} H^{Θ} = [\sum n_{p} Δ_{f} H^{Θ} (products)] - [\sum n_{r} Δ_{f} H^{Θ} (reactants)]

💡 Teacher's Secret Hint

Remember that elements in their standard states have an enthalpy of formation of zero. Although not applicable here, it's a common point to remember.

Step 3: Calculate Sum of Products' Enthalpies○ Expand

For the products (B $_{2}$ O(g) and AO $_{2}$ (g)):

\sum n_{p} Δ_{f} H^{Θ} (products) = (1 \times Δ_{f} H^{Θ} (B_{2} O(g))) + (3 \times Δ_{f} H^{Θ} ({AO}_{2} (g)))

= (1 \times + 81) + (3 \times - 393)

= 81 - 1179

= - 1098 {kJ mol}^{- 1}

💡 Teacher's Secret Hint

Be careful with the signs when multiplying. A positive stoichiometric coefficient multiplied by a negative enthalpy value results in a negative term.

Step 4: Calculate Sum of Reactants' Enthalpies○ Expand

For the reactants (B $_{2}$ O $_{4}$ (g) and AO(g)):

\sum n_{r} Δ_{f} H^{Θ} (reactants) = (1 \times Δ_{f} H^{Θ} (B_{2} O_{4} (g))) + (3 \times Δ_{f} H^{Θ} (AO(g)))

= (1 \times + 9.7) + (3 \times - 110)

= 9.7 - 330

= - 320.3 {kJ mol}^{- 1}

💡 Teacher's Secret Hint

Ensure all given values are used with their correct stoichiometric coefficients. Double-check your arithmetic, especially when dealing with multiple terms.

Step 5: Calculate Overall Enthalpy of Reaction○ Expand

Now, substitute the calculated sums into Hess's Law equation:

Δ_{r} H^{Θ} = (Sum of products' enthalpies) - (Sum of reactants' enthalpies)

= (- 1098) - (- 320.3)

= - 1098 + 320.3

= - 777.7 {kJ mol}^{- 1}

The standard enthalpy change for the reaction is $- 777.7 {kJ mol}^{- 1}$ .

💡 Teacher's Secret Hint

A common mistake is forgetting to correctly handle the double negative sign when subtracting a negative sum. $(- A) - (- B) = - A + B$ .

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Chemistry Question 128 – AP-EAMCET 2026

Clue 1 of 3: Enthalpy of Reaction from Enthalpies of Formation

🧠 Full Solution Path