For a certain chemical reaction, the standard Gibbs free energy of reaction is -67.5 kJ. Calculate the temperature at which the equilibrium constant [tex]K=2.8 \times 10^{12}[/tex]. Round your answer to the nearest degree.
Answer (1)
Use the formula Δ G ∘ = − RT ln K .
Rearrange to solve for temperature: T = − R l n K Δ G ∘ .
Substitute the given values: T = − 8.314 × l n ( 2.8 × 1 0 12 ) − 67500 .
Calculate and round to the nearest degree: T = 283 K .
Explanation
Problem Setup We are given the standard Gibbs free energy of a chemical reaction, Δ G ∘ = − 67.5 kJ = − 67500 J , and the equilibrium constant, K = 2.8 × 1 0 12 . We want to find the temperature T at which this equilibrium occurs.
Relevant Equation We will use the relationship between the standard Gibbs free energy, the equilibrium constant, and the temperature: Δ G ∘ = − RT ln K where R is the ideal gas constant, R = 8.314 J/(mol K) .
Solving for Temperature We need to rearrange the equation to solve for T : T = − R ln K Δ G ∘
Substitution Now, we substitute the given values into the equation: T = − 8.314 J/(mol K) × ln ( 2.8 × 1 0 12 ) − 67500 J
Calculation Calculating the value of T : T = 8.314 × ln ( 2.8 × 1 0 12 ) 67500 ≈ 8.314 × 28.66 67500 ≈ 238.29 67500 ≈ 283.26 K
Final Answer Rounding the temperature to the nearest degree, we get T = 283 K .
Examples
Understanding the relationship between Gibbs free energy, equilibrium constant, and temperature is crucial in various chemical processes. For example, in designing industrial chemical reactors, it's essential to know the temperature at which a reaction will reach a desired equilibrium state. This calculation helps optimize reaction conditions to maximize product yield and minimize energy consumption, leading to more efficient and cost-effective chemical production.
