The KC for the reaction
I2 --> 2I
is 3.8 x 10 ^ -5 at 727 degrees C.
Calculate Kc and Kp for the equilibrium of
2I -> I2
at the same conditions.
I don't understand how to set up or solve this problem. I know Kc=[I]^2 / [I2] = 3.8x10^-5 for the first equation but I don't understand how to relate it to the second equation or how to use the temperature in the process?
Oops, my problem states I2 is a gas. I forgot to include it. so does this change the equation?
To relate the equilibrium constant (Kc or Kp) for the first equation to the second equation, you need to use the concept of the reverse reaction. The second equation is the reverse reaction of the first equation.
For the first equation:
I2 ⇌ 2I
The equilibrium constant expression is given by:
Kc = [I]^2 / [I2]
Now, let's relate it to the second equation (reverse reaction):
2I ⇌ I2
To obtain the equilibrium constant expression for the reverse reaction, you need to invert the equation and the corresponding equilibrium constant:
(I2 ⇌ 2I) --> reverse reaction
(I2 ⇌ 2I)⁻¹ --> reversed equation
Invert the equilibrium constant as well:
(1 / Kc) --> inverse equilibrium constant
Now, the equilibrium constant expression for the reverse reaction becomes:
Kc(reverse) = 1 / Kc = 1 / ([I]^2 / [I2])
To calculate Kp, the equilibrium constant in terms of partial pressures, you need to apply the ideal gas law and the concept of gas stoichiometry.
Assuming the reaction is taking place in the gas phase, using the ideal gas law, you can write:
PI2 = 2 * PI
Where PI2 is the partial pressure of I2 and PI is the partial pressure of I.
You can substitute these relationships into the equilibrium constant expression to obtain Kp.
Now, to incorporate the temperature, you need to consider the Van't Hoff equation. The equation is given by:
ln(K2/K1) = (ΔH°/R) * (1/T1 - 1/T2)
Where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively. ΔH° is the enthalpy change, and R is the gas constant.
Since you are given the Kc at a certain temperature (727 degrees C), you can use this equation to determine the equilibrium constant at the same temperature.
Remember to convert the temperature to Kelvin by adding 273 to the Celsius temperature before substituting it into the equation.
By following these steps, you should be able to calculate the Kc and Kp for the equilibrium of 2I ⇌ I2 at the same conditions.
Kp = Kc(RT)^delta n.
BTW, isn't I2 a solid. Then does I2 go in the denominator.
Yes and no.
You have Kc given in the problem. All you need to do is to substitute Kc into the Kp = Kc*(RT)^delta n and solve for Kp and you're done. The only thing changing I2 from solid to gas changes is delta n. For I2(g) ==>2I^-(g) delta n will be 2-1 = 1