What is Δr Gominus for the following reaction at 298 K? 2 X(g) + Y(g) → 2 Z(g) (Δ Uominus = -10.5 kJ; Δ Sominus = -44 JK-1; R = 8.3 JK-1mol-1)

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q129 – AP-EAMCET 2026

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Home Chemistry Chemical ThermodynamicsQuestion #292

What is

Δ_{r} G^{⊖}

for the following reaction at

298 K

2 X (g) + Y (g) \to 2 Z (g)

(

Δ U^{⊖} = - 10.5 kJ

;

Δ S^{⊖} = - 44 {JK}^{- 1}

;

R = 8.3 {JK}^{- 1} {mol}^{- 1}

)

Clue 1 of 3: Thermodynamic Relationship

The change in Gibbs free energy ( $Δ G$ ) is related to the change in enthalpy ( $Δ H$ ) and entropy ( $Δ S$ ) by the equation $Δ G = Δ H - T Δ S$ . However, the internal energy ( $Δ U$ ) is given, so we need to first find $Δ H$ .

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Step 1: Identify Given Values and Determine

Δ n_{g}

✦ Active

The problem asks for the standard Gibbs free energy change ( $Δ_{r} G^{⊖}$ ) for the reaction at $298 K$ . We are given the standard internal energy change ( $Δ U^{⊖}$ ), standard entropy change ( $Δ S^{⊖}$ ), and the gas constant ( $R$ ). The reaction is: $2 X (g) + Y (g) \to 2 Z (g)$ From the reaction, we can calculate the change in the number of moles of gaseous species ( $Δ n_{g}$ ):

Δ n_{g} = (moles of gaseous products) - (moles of gaseous reactants)

Δ n_{g} = (2) - (2 + 1) = 2 - 3 = - 1 mol

Given values: $Δ U^{⊖} = - 10.5 kJ = - 10500 J$ $Δ S^{⊖} = - 44 {JK}^{- 1} {mol}^{- 1}$ $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ $T = 298 K$

💡 Teacher's Secret Hint

Always convert all energy values to a consistent unit (e.g., Joules) at the beginning of the calculation to avoid errors later on.

Step 2: Calculate Standard Enthalpy Change (

Δ H^{⊖}

)○ Expand

The relationship between standard enthalpy change ( $Δ H^{⊖}$ ) and standard internal energy change ( $Δ U^{⊖}$ ) for a reaction involving gases is:

Δ H^{⊖} = Δ U^{⊖} + Δ n_{g} R T

Substitute the known values into the equation:

Δ H^{⊖} = - 10500 J + (- 1 mol) (8.3 {JK}^{- 1} {mol}^{- 1}) (298 K)

Δ H^{⊖} = - 10500 J - 2473.4 J

Δ H^{⊖} = - 12973.4 J

💡 Teacher's Secret Hint

Pay close attention to the sign of $Δ n_{g}$ . A negative value means the number of gas moles decreases, often leading to a more negative $Δ H$ when $Δ U$ is already negative, as work is done by the surroundings on the system.

Step 3: Calculate Standard Gibbs Free Energy Change (

Δ G^{⊖}

)○ Expand

Now, we can calculate the standard Gibbs free energy change ( $Δ G^{⊖}$ ) using the standard enthalpy change ( $Δ H^{⊖}$ ), standard entropy change ( $Δ S^{⊖}$ ), and temperature ( $T$ ):

Δ G^{⊖} = Δ H^{⊖} - T Δ S^{⊖}

Substitute the calculated $Δ H^{⊖}$ and given $Δ S^{⊖}$ and $T$ :

Δ G^{⊖} = - 12973.4 J - (298 K) (- 44 {JK}^{- 1} {mol}^{- 1})

Δ G^{⊖} = - 12973.4 J - (- 13112 J)

Δ G^{⊖} = - 12973.4 J + 13112 J

Δ G^{⊖} = 138.6 J

Rounding the result to one decimal place, we get $138.6 J$ . Comparing this with the given options, $139.0 J$ is the closest value.

💡 Teacher's Secret Hint

Double-check the signs when subtracting $T Δ S^{⊖}$ . If $Δ S^{⊖}$ is negative, then $- T Δ S^{⊖}$ will be positive, as seen in this calculation.

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Chemistry Question 129 – AP-EAMCET 2026

Clue 1 of 3: Thermodynamic Relationship

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