Chemistry Q129 – AP-EAMCET 2026

Home Chemistry Chemical ThermodynamicsQuestion #294

17^{\circ} C

, the enthalpy change and entropy change of the reaction given below are respectively

- 12.55 {kJ mol}^{- 1}

and

5.0 {JK}^{- 1}

Zn (s) + 2 HCl (a q) \to {ZnCl}_{2} (a q) + H_{2} (g)

What is

Δ G

(in

{kJ mol}^{- 1}

) for this reaction?

Clue 1 of 3: Gibbs Free Energy

The change in Gibbs free energy ( $Δ G$ ) determines the spontaneity of a chemical reaction at constant temperature and pressure.

🧠 Full Solution Path

Step 1: Identify Given Values and Equation✦ Active

The problem provides the enthalpy change ( $Δ H$ ), entropy change ( $Δ S$ ), and temperature ( $T$ ) for a reaction. We need to calculate the Gibbs free energy change ( $Δ G$ ). The relevant equation is the Gibbs-Helmholtz equation:

Δ G = Δ H - T Δ S

Given values are: $Δ H = - 12.55 {kJ mol}^{- 1}$ $Δ S = 5.0 {JK}^{- 1}$ $T = 17^{\circ} C$

💡 Teacher's Secret Hint

Always list out the given quantities and the target variable before starting calculations. This helps in identifying necessary formulas and unit conversions.

Step 2: Convert Units to Ensure Consistency○ Expand

For the Gibbs-Helmholtz equation, temperature must be in Kelvin, and the units of $Δ H$ and $T Δ S$ must be consistent. Since $Δ H$ is in $kJ$ , $Δ S$ should also be converted to ${kJ K}^{- 1}$ .

Convert temperature from Celsius to Kelvin:

T (K) = T (^{\circ} C) + 273.15

$T = 17 + 273.15 = 290.15 K$

Convert entropy change from ${J K}^{- 1}$ to ${kJ K}^{- 1}$ :

Δ S = 5.0 {JK}^{- 1} \times \frac{1 kJ}{1000 J} = 0.0050 {kJ K}^{- 1}

💡 Teacher's Secret Hint

A common error is forgetting to convert temperature to Kelvin or ensuring that the energy units (Joules vs. kiloJoules) are consistent across all terms.

Step 3: Calculate Gibbs Free Energy Change○ Expand

Now, substitute the converted values into the Gibbs-Helmholtz equation:

Δ G = Δ H - T Δ S

Substitute the values:

Δ G = - 12.55 {kJ mol}^{- 1} - (290.15 K) (0.0050 {kJ K}^{- 1} {mol}^{- 1})

Perform the multiplication:

T Δ S = 290.15 \times 0.0050 = 1.45075 {kJ mol}^{- 1}

Complete the subtraction:

Δ G = - 12.55 - 1.45075 = - 14.00075 {kJ mol}^{- 1}

Rounding to two decimal places, or matching the options provided, $Δ G \approx - 14.00 {kJ mol}^{- 1}$ .

💡 Teacher's Secret Hint

Pay close attention to the sign convention. A negative $Δ G$ indicates a spontaneous reaction under the given conditions.

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Chemistry Question 129 – AP-EAMCET 2026

Clue 1 of 3: Gibbs Free Energy

🧠 Full Solution Path