Determine the delta G(o) and the equilibrium constant for the following reaction at 323 K.
2Br- + I2 --> Br2 + 2I-
I got 100 kj/mol
and K 7.19x10^16
but not sure if this is right
To determine the delta G(o) and the equilibrium constant (K) for the given reaction at 323 K, you need to use the thermodynamic equations related to free energy (delta G) and equilibrium constant (K).
The standard free energy change (delta G(o)) of a reaction can be calculated using the equation:
delta G(o) = -RT ln(K)
Where:
- delta G(o) is the standard free energy change
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- K is the equilibrium constant
In this case, the temperature given is 323 K.
First, let's calculate the delta G(o):
delta G(o) = -RT ln(K)
delta G(o) = - (8.314 J/(mol·K)) * (323 K) * ln(7.19x10^16)
Note: The given value of 100 kJ/mol does not seem to be accurate, as delta G(o) is typically reported in J/mol, not kJ/mol.
Calculating the value of delta G(o) using the equation above will yield the correct result.
Next, to determine the equilibrium constant (K) from delta G(o), we use the equation:
delta G(o) = -RT ln(K)
Rearranging the equation and solving for K:
ln(K) = -delta G(o) / (RT)
K = e^(-delta G(o) / (RT))
Plug in the calculated value of delta G(o) and solve for K using the given temperature (323 K).
Once you have calculated the correct values for delta G(o) and K, you can compare them to the values you initially obtained to check if they were correct.