Question

At 700 oC, Kc = 20.4 for the reaction SO2 (g) + ½ O2 (g)  SO3 (g)

Answer 1

a) What is the value of Kc for the reaction of SO3 (g)  SO2 (g) + ½ O2 (g)?





b) What is the value of Kc for the reaction 2 SO2 (g) + O2 (g)  2 SO3 (g)?






c) What is the value of Kp for the reaction 2 SO2 (g) + O2 (g)  2 SO3 (g)

Answer 2

Well, I assume here we're given Kc and the temperature for good reason. Kc/Kp,K,Ksp any equilibrium constant are temperature dependent. When we raise or lower the temperature we'll cause the equilibrium to shfit which might result in a different Kc value, but sometimes it's not that clear though raising the temperature generally increases the rate of our reaction. but it depends also whether our reaction is exothermic or endothermic.

The formula we need is below:

where delta n is the change in the number of moles {products-reactants}



\[K_{p} =K_{c}RT^{\delta N} \]

Answer 3

what value do i use to find the difference in products and reactants

Answer 4

difference in the number of moles for your reaction. Remember T = Temperature in Kelvins, R is the gas constant = 0.082.