Answer :
Final Answer:
The equilibrium constant (K) for the given reaction at 277.0 K is
1.59 x [tex]10^7[/tex].
Explanation:
To calculate the equilibrium constant (K) for the given reaction, we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG = Gibbs Free Energy Change
ΔH = Enthalpy Change
T = Temperature in Kelvin
ΔS = Entropy Change
First, we need to convert the enthalpy change (ΔH) from kilojoules to joules:
ΔH = 46.5 , kJ = 46,500 , J
Given that ΔS is -124.8 J/K, and the temperature (T) is 277.0 K, we can substitute these values into the equation to find ΔG:
ΔG = 46,500 , J - (277.0 , K)(-124.8 , J/K) = 79,604 , J]
Now, we can use ΔG to calculate the equilibrium constant (K) using the following equation:
[ΔG = -RTln(K)]
Where:
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
K = Equilibrium constant
Rearranging the equation to solve for K:
[tex]\[K = e^{\frac{-ΔG}{RT}}\][/tex]
Substitute the values:
[tex]\[K = e^{\frac{-79,604 \, J}{(8.314 \, J/(mol·K))(277.0 \, K)}}\][/tex]
Calculating this expression gives:
[tex]\[K \approx 1.59 \times 10^7\][/tex]
So, the equilibrium constant (K) for the given reaction at 277.0 K is approximately 1.59 x [tex]10^7[/tex].
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