Answer :
The question is incomplete, the complete question is:
For a certain chemical reaction, the standard Gibbs free energy of reaction is 144. kJ. Calculate the temperature at which the equilibrium constant K = [tex]5.9\times 10^{-26}[/tex] .
Round your answer to the nearest degree.
Answer:
25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].
Explanation:
[tex]\Delta G^o=-RT\ln K[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = 144.0 kJ=144,000 J (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314 J/K mol[/tex]
T = temperature at which reaction is occurring = ?
K = Equilibrium constant of the reaction =[tex]5.9\times 10^{-26}[/tex]
Putting values in above equation, we get:
[tex]144,000 J/mol=-(8.3145J/Kmol)\times T\times \ln [5.9\times 10^{-26}][/tex]
[tex]T=\frac{144,000 J/mol}{-(8.314 J/Kmol)\times \ln [5.9\times 10^{-26}]}[/tex]
T = 298.15 K
T = 298.15 - 273 °C = 25°C
25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].
The temperature at which the equilibrium constant K = 5.9 × 10⁻²⁶ is 25 degree C.
How we calculate the temperature for the given question?
We can calculate the temperature for the given question as follow:
ΔG⁰ = -RT ln K, where
ΔG⁰ = standard Gibbs free energy = 144.0 kJ = 1,44,000 J (given)
R = gas constant = 8.314 J/mole K
T = temperature = to find ?
K = equilibrium constant = 5.9 × 10⁻²⁶
Putting all these values in the above equation, we get
1,44,000 J/mole = -(8.314 J/mole K) × (T) × ln(5.9 × 10⁻²⁶)
T = - (1,44,000 / 8.314 × ln 5.9× 10⁻²⁶ )
T = 298.15 K
Or T = 298.15 - 273 °C = 25°C
Hence, 25 degree C is the temperature at which the equilibrium constant K = 5.9 × 10⁻²⁶.
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