SO2(g) + NO2(g) SO3(g) + NO (g)
At a given temperature, analysis of an equilibrium mixture found [SO2] = 4.00 M, [NO2] = 0.500 M, [SO3] = 3.00 M, and [NO] = 2.00 M.
(a) What is the new equilibrium concentration of NO when 1.50 moles of NO2 are added to the equilibrium mixture? Assume a 1.00 L container.
[NO] = 2.84M <<answer
(b) How many moles/liter of NO2 would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO3 from 3.00 M to 4.10 M at the same temperature?
[NO2] =?????????????/
i need help on part b how do u do this? thanks!!
K = [SO3] [NO] / [SO2] [NO2]
K = (0.6)(2.0) / (0.3)(0.1) = 40
At the new equilibrium,
[SO2] = 0.3 + [SO2]add - x = 0.8 M ;
[NO2] = 0.1-x ; [NO] = 2.0+x ; [SO3] = 0.6+x
K = [0.6+x] [2.0+x] / ( [0.8] [0.1-x] ) = 40
x = 0.0577 M
[SO2] = 0.8 M ; [NO2]=0.1-x = 0.0423 M ;
[SO3] = 0.6+x = 0.6577 M ; [NO] = 2.0+x = 2.0577 M
Verif. → (0.6577*2.0577) / (0.8*0.0423) = 39.993
I tried to answer the hwk using this tip for part B, but I didn't get it right...is there another way to set it up?
SO2(g) + NO2(g) SO3(g) + NO (g)
At a given temperature, analysis of an equilibrium mixture found [SO2] = 0.300M, [NO2] = 0.100 M, [SO3] = 0.600 M, and [NO] = 2.00 M. At the same temp., extra SO2 was added to make SO2 is added to make [SO2}= 0.800M. Calculate the composition of the mixture when equilibrium has been reestablished
SO2(g) + NO2(g) ⇄ SO3(g) + NO(g)
had reached a state of equilibrium, was found to contain
0.40 mol L-1 SO3 , and 0.30 mol L-1 NO,
0.15 mol L-1NO2 , and 0.20 mol L-1 SO2.
Why did the chicken cross the road? To get to the other side... of the equilibrium!
Now, let's tackle part B of your question. To find out how many moles/liter of NO2 you need to add to the original equilibrium mixture, we can use the equilibrium expression for the reaction:
Kc = ([SO3] * [NO]) / ([SO2] * [NO2])
We are given the following equilibrium concentrations:
[SO3] = 3.00 M
[NO] = 2.00 M
[SO2] = 4.00 M
We know that Kc does not change with temperature, so we can use it to find the new equilibrium concentration of NO2 by setting up an equation:
4.10 = (3.00 + x) * (2.00 + x) / (4.00 * x)
Now, put on your equation-solving hat and find the value of x, which represents the change in concentration of NO2. Once you have x, you can add it to the initial concentration of [NO2] = 0.500 M to find the new equilibrium concentration.
I hope this helps! Let me know if you need any more assistance.
To solve part B of this question, you need to use the concept of the equilibrium constant and the reaction stoichiometry. The equilibrium constant, denoted as K, is a value that relates the concentrations of reactants and products at equilibrium.
The balanced equation for the given reaction is:
SO2(g) + NO2(g) SO3(g) + NO(g)
First, we need to find the equilibrium constant expression for this reaction. The equilibrium constant expression is written using the concentrations of the substances involved:
K = [SO3] / ([SO2] * [NO2])
Given that [SO2] = 4.00 M, [NO2] = 0.500 M, and [SO3] = 3.00 M, we can calculate the initial value of K using these concentrations:
K_initial = [3.00] / ([4.00] * [0.500])
Now, we want to increase the equilibrium concentration of SO3 from 3.00 M to 4.10 M. Let's assume that x is the change in the concentration of [NO2] required to reach this new equilibrium concentration.
So, the new concentrations will be:
[SO2] = 4.00 M
[NO2] = 0.500 M + x (since we are adding x moles)
[SO3] = 4.10 M
[NO] = unknown (let's call it y)
To find the value of x, we need to use the fact that K remains constant at equilibrium. Therefore, we can set up the following equation:
K_initial = (4.10) / (4.00 * (0.500 + x))
Now, we can solve for x:
[3.00] / ([4.00] * [0.500]) = (4.10) / (4.00 * (0.500 + x))
Solving this equation will give you the value of x, which represents the change in the concentration of [NO2]. Once you find x, you can add it to the initial concentration of [NO2] (0.500 M) to get the new equilibrium concentration, [NO2].
K = [SO3][NO]/[SO2][NO2]
K = (3.0)(2.0)/(4.0)(0.5)= 3.0
(a) Some of the added NO2 will be used up causing a shift to the right. Let that be equal to x.
At the new equilibrium,
[SO2]=4.0-x, [NO2]=0.5+1.5-x=4.0-x
[NO]=2.0+x, [SO3]=3.0+x,
Set up:
K = [3.0+x][2.0+x]/[4.0-x][2.0-x]
Simplify and solve for x
[NO]=2.0+x
(b)
Let the moles of NO2 added = x
Set up:
3.0 = [3.0+1.1][2.0+1.1]/[4.0-1.1][0.5+x]
Solve for x