An equilibrium reaction which has K1 = 6.5 x 10¯2 at 100oC and has ∆Horxn = -49.5 kJmol¯1. What is the value of K at 250 oC?
I can't get the right answer, which is supposed to be 6.68 x 10^-4.
How does one do this? Thanks
Should be 100 degrees C and 250 degrees C
ln(K1/K2)=(∆Horxn/R)((1/T2)-(1/T1))
ln((6.5x10^-2)/K2)=((-49.5 kJmol^-1)/R)((1/(250+273.15))-(1/(100+273.15)))
That's it! Just solve for K2. Don't forget that your R must have the units of kJ/molK.
Hope this helps. Cheers!
To find the value of K at 250 oC, you can use the Van't Hoff equation. The Van't Hoff equation relates the equilibrium constant (K) of a reaction with temperature.
The Van't Hoff equation is expressed as:
ln(K2/K1) = (∆H/R) * (1/T1 - 1/T2)
Where:
K1 = equilibrium constant at temperature T1
K2 = equilibrium constant at temperature T2
∆H = enthalpy change of the reaction
R = gas constant (8.314 J/mol•K)
T1 = initial temperature (in Kelvin)
T2 = final temperature (in Kelvin)
Now, let's plug in the given values into the equation:
K1 = 6.5 x 10¯^2
∆H = -49.5 kJ/mol (Note: make sure to convert to J/mol, which is -49,500 J/mol)
R = 8.314 J/mol•K
T1 = 100 oC (convert to Kelvin, so it becomes 100 + 273 = 373 K)
T2 = 250 oC (convert to Kelvin, so it becomes 250 + 273 = 523 K)
Now we can calculate ln(K2/K1):
ln(K2/6.5 x 10¯^2) = (-49,500 J/mol / 8.314 J/mol•K) * (1/373 K - 1/523 K)
Simplifying further:
ln(K2/6.5 x 10¯^2) = -5970.8 * (0.00268 - 0.00191)
ln(K2/6.5 x 10¯^2) = -5970.8 * 0.00077
ln(K2/6.5 x 10¯^2) = -4.6049
Next, we can solve for K2:
K2/6.5 x 10¯^2 = e^(-4.6049)
K2 = (e^(-4.6049)) * (6.5 x 10¯^2)
Calculating this result gives K2 ≈ 6.68 x 10¯^4.
Therefore, the value of K at 250 oC is approximately 6.68 x 10¯^4.