Question

For a certain chemical reaction, the standard Gibbs free energy of reaction at 10.0 °C is 149. kJ. Calculate the equilibrium constant K for this reaction to 2 significant digits. Round your answer K ?

Answer 1

To calculate the equilibrium constant K for a chemical reaction, we use the standard Gibbs free energy of reaction. For the given ΔG° of 149 kJ at 10.0 °C, we convert it to joules and then calculate K using the formula K = exp(-ΔG° / (R * T)). The calculated value of K is ≈ 2.42 × 10⁻³.

Step-by-step explanation:

To calculate the equilibrium constant K for a chemical reaction, we can use the standard Gibbs free energy of reaction ΔG°. To do this, we need to convert the given ΔG° from kilojoules to joules and then divide it by the gas constant R (8.314 J/mol K) multiplied by the temperature T in Kelvin. So, for the given ΔG° of 149 kJ at 10.0 °C, we have:

1. Conversion of ΔG° to joules: ΔG° = 149 kJ * 1000 J/1 kJ = 149,000 J
2. Calculation of equilibrium constant: K = exp(-ΔG° / (R * T))
3. Substituting values: K = exp(-149,000 J / (8.314 J/mol K * (10.0 + 273.15) K))
4. Calculating K: K ≈ 2.42 × 10⁻³

Answer 2

Answer: The equilibrium constant for this reaction is
`3.1* 10^(-28)`

Step-by-step explanation:

To calculate the equilibrium constant (at 15°C) for given value of Gibbs free energy, we use the relation:

`\Delta G^o=-RT\ln K_(eq)`

where,

`\Delta G^o` = standard Gibbs free energy = 149. kJ/mol = 149000 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant =
`8.314J/K mol`

T = temperature =
`15^oC=[273+15]K=283K`

`K_(eq)` = equilibrium constant at 10°C = ?

Putting values in above equation, we get:

`149000J/mol=-(8.314J/Kmol)* 283K* \ln K_(eq)\\\\K_(eq)=3.1* 10^(-28)`

Hence, the equilibrium constant for this reaction is
`3.1* 10^(-28)`