the standard free energy change for a chemical reaction is +13.3 kj/ mol. What is the equilibrium constant for the reaction at 125 degree C ?
(R= 8.314 j/mol )
Keq= -e^(-Gibbsfree/RT)
To determine the equilibrium constant (K) for a chemical reaction given the standard free energy change (ΔG°), we use the following equation:
ΔG° = -RT ln(K)
Where:
ΔG° = standard free energy change
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin
K = equilibrium constant
Given:
ΔG° = +13.3 kJ/mol
T = 125 °C = 125 + 273.15 = 398.15 K
R = 8.314 J/mol·K
Let's substitute these values into the equation and solve for K:
ΔG° = -RT ln(K)
13.3 kJ/mol = -(8.314 J/mol·K) × (398.15 K) ln(K)
Converting kJ to J:
13.3 × 10^3 J/mol = -(8.314 J/mol·K) × (398.15 K) ln(K)
Now, we can rearrange the equation to solve for ln(K):
ln(K) = -13.3 × 10^3 J/mol / [(8.314 J/mol·K) × (398.15 K)]
ln(K) = -39.70
Using the natural logarithm (ln) property, we can convert ln(K) to K:
K = e^(-39.70)
Calculating K using a calculator:
K ≈ 2.6 × 10^(-18)
Therefore, the equilibrium constant for the reaction at 125 °C is approximately 2.6 × 10^(-18).
To calculate the equilibrium constant (K) for a chemical reaction using the standard free energy change (∆G°), you can use the following equation:
∆G° = -RT * ln(K)
Where:
∆G° is the standard free energy change (in joules/mol)
R is the gas constant (8.314 J/mol*K)
T is the temperature in Kelvin
Since the temperature given is in degrees Celsius (°C), you need to convert it to Kelvin by adding 273.15:
T = 125°C + 273.15 = 398.15 K
Now, substitute the values into the equation:
∆G° = +13.3 kJ/mol * 1000 J/1 kJ = +13,300 J/mol
R = 8.314 J/mol*K
T = 398.15 K
∆G° = -RT * ln(K)
13,300 J/mol = -8.314 J/mol*K * 398.15 K * ln(K)
Now, we can solve for ln(K):
ln(K) = -13,300 J/mol / (-8.314 J/mol*K * 398.15 K)
ln(K) = 4.015
To find K, take the exponential of both sides of the equation:
K = e^(ln(K))
K = e^(4.015)
Using a calculator or software, calculate the value of e^(4.015), which is equal to approximately 55.40.
Therefore, the equilibrium constant (K) for the reaction at 125°C is approximately 55.40.