A chemist fills a reaction vessel with 0.536 g aluminum hydroxide $\left( Al ( OH )_3\right)$ solid, 0.337 M aluminum $\left( Al ^{3+}\right)$ aqueous solution, and 0.196 M hydroxide $\left( OH ^{-}\right)$aqueous solution at a temperature of $25.0^{\circ} C$. Under these conditions, calculate the reaction free energy $\Delta G$ for the following chemical reaction: $Al(OH)_3(s) \rightharpoonup Al^{3+}(aq)+3 OH^{-}(aq)$ Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Determine the standard Gibbs free energy change ( Δ G ∘ ) using the given standard Gibbs free energies of formation.
Calculate the reaction quotient (Q) using the given concentrations of A l 3 + and O H − .
Use the Gibbs free energy equation, Δ G = Δ G ∘ + RT ln Q , to calculate the reaction free energy ( Δ G ) under non-standard conditions.
Convert the result to kilojoules and round to the nearest kilojoule, resulting in 176 ​ kJ/mol.

Explanation

Find standard Gibbs free energies First, we need to find the standard Gibbs free energy of formation ( Δ G f ∘ ​ ) for each species in the reaction from the ALEKS Data tab. We have:


Δ G f ∘ ​ ( A l 3 + ( a q )) = − 489 kJ/mol
Δ G f ∘ ​ ( O H − ( a q )) = − 157 kJ/mol
Δ G f ∘ ​ ( A l ( O H ) 3 ​ ( s )) = − 1151 kJ/mol


Calculate standard free energy change Next, calculate the standard free energy change ( Δ G ∘ ) for the reaction using the formula: Δ G ∘ = ∑ Δ G f ∘ ​ ( products ) − ∑ Δ G f ∘ ​ ( reactants )

In this case: Δ G ∘ = [ Δ G f ∘ ​ ( A l 3 + ( a q )) + 3 × Δ G f ∘ ​ ( O H − ( a q ))] − [ Δ G f ∘ ​ ( A l ( O H ) 3 ​ ( s ))] Δ G ∘ = [ − 489 + 3 × ( − 157 )] − [ − 1151 ] Δ G ∘ = [ − 489 − 471 ] + 1151 Δ G ∘ = − 960 + 1151 Δ G ∘ = 191 kJ/mol

Calculate the reaction quotient Now, calculate the reaction quotient (Q) using the given concentrations: Q = [ A l 3 + ] [ O H − ] 3 Q = ( 0.337 ) ( 0.196 ) 3 Q = ( 0.337 ) ( 0.007529 ) ≈ 0.002537

Convert temperature to Kelvin Convert the temperature from Celsius to Kelvin: T ( K ) = T ( ∘ C ) + 273.15 T = 25.0 + 273.15 = 298.15 K

Calculate the reaction free energy Use the Gibbs free energy equation to calculate the reaction free energy ( Δ G ) under the given non-standard conditions: Δ G = Δ G ∘ + RT ln Q Where R is the ideal gas constant (8.314 J/(mol*K)). First, convert Δ G ∘ to J/mol: Δ G ∘ = 191 kJ/mol = 191000 J/mol Now, substitute the values into the equation: Δ G = 191000 + ( 8.314 ) ( 298.15 ) ln ( 0.002537 ) Δ G = 191000 + ( 8.314 ) ( 298.15 ) ( − 5.974 ) Δ G = 191000 − 14790.4 ≈ 176209.6 J/mol

Convert to kJ/mol and round Convert the result from joules to kilojoules by dividing by 1000: Δ G ( k J ) = 1000 Δ G ( J ) ​ Δ G ( k J ) = 1000 176209.6 ​ ≈ 176.2 kJ/mol Round the final answer to the nearest kilojoule: Δ G ≈ 176 kJ/mol

Final Answer Therefore, the reaction free energy Δ G for the given chemical reaction under the specified conditions is approximately 176 kJ/mol.


Examples
Understanding Gibbs free energy is crucial in various real-world applications, such as designing efficient chemical processes and predicting reaction spontaneity. For instance, in the development of new pharmaceuticals, chemists use Gibbs free energy calculations to determine whether a drug molecule will bind spontaneously to its target protein in the body. Similarly, environmental scientists use these calculations to assess the feasibility of pollutant degradation reactions in contaminated soil or water. By manipulating reaction conditions to achieve a negative Gibbs free energy, scientists can drive reactions toward desired outcomes, optimizing processes and minimizing waste.

Answered by GinnyAnswer | 2025-07-06

To find the reaction free energy Δ G for the reaction of aluminum hydroxide, we calculated its standard free energy change using Gibbs free energy of formation values. We then computed the reaction quotient based on the given concentrations and used the Gibbs free energy equation to find Δ G , resulting in an approximate value of 176 kJ/mol. Therefore, the final answer is 176 kJ/mol .
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Answered by Anonymous | 2025-07-17