Question

For the following reaction, Kp = 1.45 x 10^-5 at 500C. What is the value of Kc?

Answer 1

A) 5.83*10^-2

Step-by-step explanation:

The information given from the exercise is:

- Kp value: 1.45x10^-5

- Temperature (T): 500°C

- Chemical reaction

We can calculate the value of Kc by replacing the values of R (gas constant: 0.082 atm*L/mol*K), temperature (T) and Δn in the Kp formula:

`K_p=K_c*(R*T)^(\Delta n)`

In this formula, Δn is the result of the difference between the products coefficients and the reactants ccoefficents:

`\begin{gathered} \Delta n=n_(products)-n_(reactants) \\ \Delta n=2-4 \\ \Delta n=-2 \end{gathered}`

It is important to convert the unit of temperature from °C to K:

`500\text{ + }273=773K`

Now we can replace the values:

`\begin{gathered} K_p=K_c*(R*T)^(\Delta n) \\ 1.45*10^(-5)=K_c*(0.082*773)^((-2)) \\ 1.45*10^(-5)=K_c*(2.48*10^(-4)) \\ (1.45*10^(-5))/(2.48*10^(-4))=K_c \\ 5.83*10^(-2)=K_c \end{gathered}`

So, the value of Kc is 5.83*10^-2.