hi
i have difficulty in this question.any help would be much appreciated.
1) a mixture of 0.0080 mol of So2(g) and 0.0056mol of O2(g) is placed in one liter container at 100K. when equilibrium is established, 0.0040 mol of So3(g)is present
2SO2(g)+O2(g)->2So3(g)
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a) what are the equilibrium concentration of SO2 and O2?b)what is the value of K for the equilibrium at 100k?
(a) To form 0.0040 mol of SO3, you must have used up 0.0040 moles of SO2 and 0.0020 moles of O2. That is because of the coefficients in the reaction itself.
(b) You end up with the following amounts: 0.0040 mol SO2, 0.036 mol O2 and 0.0040 mol SO3. They are still contained in the one liter container, so those are also the molar concentrations in mole/liter.
The value of Kc is [SO3]^2/[SO2]^2[O2].
Use the final values.
To solve the given question and find the equilibrium concentrations and the value of K, we can use the concept of stoichiometry, the reaction quotient Q, and the equilibrium constant K.
a) Firstly, we need to determine the change in moles for each species in the reaction. From the balanced equation, we can see that for every 2 moles of SO2 reacted, 2 moles of SO3 are produced, and 1 mole of O2 is consumed.
Let's denote the equilibrium concentration of SO2 as x, and the equilibrium concentration of O2 as y.
Initially, we have:
SO2 = 0.0080 mol
O2 = 0.0056 mol
SO3 = 0 mol
At equilibrium, the change in moles is given by:
SO2 = 0.0080 mol - 0.0040 mol (0.0040 mol of SO2 has reacted)
O2 = 0.0056 mol - 0.0020 mol (0.0040 mol of SO2 has reacted, and 0.0020 mol of O2 has been consumed)
SO3 = 0.0040 mol (produced)
The equilibrium concentrations can be written as:
SO2 = x = 0.0080 mol - 0.0040 mol = 0.0040 mol
O2 = y = 0.0056 mol - 0.0020 mol = 0.0036 mol
SO3 = 0.0040 mol
Thus, the equilibrium concentrations of SO2 and O2 are 0.0040 mol and 0.0036 mol, respectively.
b) The equilibrium constant K can be calculated using the equilibrium concentrations of each species. The formula for K is:
K = [SO3]^2 / ([SO2]^2 * [O2])
Let's calculate it using the given data:
K = (0.0040 mol)^2 / [(0.0040 mol)^2 * (0.0036 mol)]
= 0.000016 mol^2 / (0.000064 mol^3)
= 0.25 mol^(-1)
Therefore, the value of K for the equilibrium at 100K is 0.25 mol^(-1).
I hope this helps! Let me know if you have any further questions.