At a certain temperature, the equilibrium constant for the following chemical equation is 2.90. At this temperature, calculate the number of moles of NO2(g) that must be added to 2.53 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.
SO2(g)+ NO2(g) <> SO3(g) + NO(g)
I tried doing the ICE chart but there seems to be something wrong with my calculation. -_-
Ah, I see you're having trouble with your ICE chart. Well, chemistry can definitely be a bit challenging, but don't worry, I'm here to help you out, with a side of humor!
So, let's break it down. You have 2.53 moles of SO2(g) and you want to form 1.10 moles of SO3(g) at equilibrium. The equation tells us that 1 mol of SO2 reacts with 1 mol of NO2 to form 1 mol of SO3.
Now, let's assume x moles of NO2(g) are needed to react completely with the given amount of SO2(g) to form the desired 1.10 moles of SO3(g). Since the stoichiometry is 1:1 between SO2 and NO2, we can set up an equation:
2.53 mol + x mol -> 1.10 mol
To find x, we need to subtract the initial 2.53 moles from the final 1.10 moles:
x = 1.10 mol - 2.53 mol
Oh dear, it seems we have a little problem here. We can't have negative moles, that would definitely make the chemistry world go topsy-turvy! So, in this case, it simply means that no additional moles of NO2 are needed.
Why? Well, with the given amounts, it seems that you already have an excess of NO2, enough to react completely with the SO2 and form the desired amount of SO3. So, no additional NO2(g) needs to be added.
But hey, comedy is all about timing, and chemistry humor is certainly no different. So, hang in there and keep practicing those equilibrium calculations. Remember, laughter is the best reaction!
To solve this problem, we can use the ICE (initial, change, equilibrium) table to calculate the number of moles of NO2(g) that must be added.
First, let's fill in the initial row of the ICE table:
SO2(g) + NO2(g) <> SO3(g) + NO(g)
Initial: 2.53 x 0 0
The "x" represents the number of moles of NO2(g) that must be added.
Now, let's determine the change row based on the stoichiometry of the reaction. According to the balanced equation, one mole of SO2 reacts with one mole of NO2 to form one mole of SO3 and one mole of NO. Therefore, the change in moles for each species will be equal to "x":
SO2(g) + NO2(g) <> SO3(g) + NO(g)
Initial: 2.53 x 0 0
Change: -x -x +x +x
Next, we can calculate the equilibrium concentrations using the equilibrium constant expression.
The equilibrium constant (K) is given as 2.90, which is equal to the ratio of the concentrations of the product species (SO3 and NO) to the concentration of the reactant species (SO2 and NO2). Thus, we have:
K = ([SO3] * [NO]) / ([SO2] * [NO2])
Looking at our ICE table, we can express the equilibrium concentrations as follows:
K = ([SO3] * [NO]) / ([SO2] * [NO2])
2.90 = (x * x) / (2.53 * x)
Simplifying the equation:
2.90 = x^2 / 2.53
Rearranging the equation to solve for x:
x^2 = 2.90 * 2.53
x^2 = 7.337
x ≈ √7.337
x ≈ 2.71
Therefore, approximately 2.71 moles of NO2(g) must be added to 2.53 moles of SO2(g) in order to form 1.10 moles of SO3(g) at equilibrium.
To solve this problem using an ICE (Initial, Change, Equilibrium) chart, you need to follow these steps:
Step 1: Write down the balanced equation.
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)
Step 2: Construct the ICE chart. Let's assume that 'x' moles of NO2 are added to the initial 2.53 moles of SO2. Therefore, the initial moles of SO2 will be 2.53 mol, the initial moles of NO2 will be 'x' mol, and the initial moles of SO3 and NO will both be zero.
SO2(g) NO2(g) SO3(g) NO(g)
Initial 2.53 x 0 0
Change -x -x x x
Equilibrium 2.53 - x x x x
Step 3: Set up the equilibrium expression.
The equilibrium constant expression for the reaction is:
Kc = [SO3(g)]/[SO2(g)](rest of the species are excluded as they are reacted or non-reactive)
Given that the equilibrium constant (Kc) is 2.90, we have:
2.90 = (x) / (2.53 - x)
Step 4: Solve the equation for 'x'.
Multiply both sides of the equation by (2.53 - x):
2.90 * (2.53 - x) = x
Expand and simplify the equation:
7.337 - 2.9x = x
Add 2.9x to both sides:
7.337 = 3.9x
Divide both sides by 3.9:
x = 1.883 mol
Therefore, approximately 1.883 moles of NO2(g) must be added to 2.53 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.
Try this
..........SO2 + NO2 ==> SO3 + NO
I.........2.53...0.......0.....0
C..........-x....x.......1.1...1.1
E.........2.53-x..x......1.1....1.1
Total mols = 2.53-x_x_1,1+1.1 = 4.73-x
XSO2 = (2.53-x)/(4.73-x)
XNO2 = x/(4.73-x)
XSO3 = 1.1/(4.73-x)
XNO = 1.1(4.73-x)
Then partial pressures are
pSO2 = (2.53-x)/(4.73-x)]*Ptotal
pNO2 = [x/(4.73-x)]*Ptotal
pSO3 = [1.1/(4.73-x)]*Ptotal
pNO = [1.1/4.73-x)]*Ptotal
I won't finish but here is what you do.
Substitute partial pressures from above into K expression. Solve for x = mols NO2 and you have it. Don't worry about all of this extraneous stuff; it cancels and you are left with a simple quadratic to solve.