The value of KC for the interconversion of butane and isobutane is 2.5 at 25°C. If you place 0.067 mol of butane in a 0.250-L flask at 25°C and allow equilibrium to be established, what will be the concentrations of the two forms of butane?
A)[butane]
B)[isobutane]
initial concn = 0.067/0.250 = approx 0.27 but that's just an estimate.
...........A ==> B
I.......0.027...0
C.........-x....x
E......0.027-x..x
Substitute the E line into Kc expression and solve for x.
To find the concentrations of butane and isobutane at equilibrium, we need to use the balanced chemical equation for the interconversion of butane and isobutane. The equation is:
2C₄H₁₀ (g) ⇌ C₄H₁₀ (g) + C₃H₈ (g)
According to the equation, the stoichiometric ratio of butane to isobutane is 2:1.
Given:
Initial moles of butane (C₄H₁₀) = 0.067 mol
Volume of flask = 0.250 L
We can use the ideal gas equation to convert the volume to moles using the equation:
n = PV/RT
Using the ideal gas constant R = 0.0821 L·atm/mol·K and T = 25°C = 298 K:
n = (0.250 L)(1 atm)/(0.0821 L·atm/mol·K)(298 K)
n = 0.01006 mol
Since we are dealing with 0.067 mol of butane initially, the reaction will shift to the right to form isobutane.
Let x be the change in moles of butane which will react to form isobutane.
The equilibrium concentrations will be:
[butane] = (0.067 mol - x) / (0.01006 mol + x)
[isobutane] = (x) / (0.01006 mol + x)
The equilibrium constant expression KC is given by:
KC = [isobutane] / [butane]
Given that KC = 2.5, we can set up the equation:
2.5 = (x) / (0.01006 mol + x) / [(0.067 mol - x) / (0.01006 mol + x)]
Now we can solve the equation to find the value for x, which represents the change in moles of butane.
2.5 = (x) / [(0.067 mol - x)/(0.01006 mol + x)]
Cross multiplying, we get:
2.5(0.067 mol - x) = (x)(0.01006 mol + x)
Expanding the equation:
0.1675 mol - 2.5x = 0.01006x + 0.01006x²
Rearranging and combining like terms:
0.02012x² + 2.51x - 0.1675 = 0
Solving this quadratic equation using the quadratic formula, we get:
x ≈ 0.0152 mol
Substituting this value into the equilibrium concentration expressions, we find:
[butane] = (0.067 mol - 0.0152 mol) / (0.01006 mol + 0.0152 mol) ≈ 0.0437 mol/L
[isobutane] = (0.0152 mol) / (0.01006 mol + 0.0152 mol) ≈ 0.601 mol/L
Therefore, the concentrations of butane and isobutane at equilibrium are approximately:
A) [butane] = 0.0437 mol/L
B) [isobutane] = 0.601 mol/L
To determine the concentrations of butane and isobutane at equilibrium, we need to use the given value of Kc and calculate the equilibrium concentrations based on the initial amount of butane placed in the flask.
Let's use the following variables:
[x] represents the concentration of butane (in mol/L) at equilibrium.
[y] represents the concentration of isobutane (in mol/L) at equilibrium.
0.067 mol is the initial moles of butane placed in the flask.
Based on the balanced equation for interconversion of butane and isobutane:
butane ⇌ isobutane
Since the stoichiometric coefficients for both butane and isobutane are 1, we can assume that the change in concentration for both gases will be x and y, respectively, when equilibrium is attained.
At equilibrium, the concentration of butane will be (0.067 - x) since x moles of butane have reacted to form isobutane.
Using the given value of Kc (Kc = 2.5), we can set up the equilibrium expression:
Kc = [isobutane] / [butane]
Substituting the equilibrium concentrations, we have:
2.5 = y / (0.067 - x)
Simplifying the expression, we have:
2.5(0.067 - x) = y
Since the total initial volume of the flask is 0.250 L, the sum of the equilibrium concentrations should equal 0.067 mol / 0.250 L = 0.268 M:
[x] + [y] = 0.268
Now we have two equations:
2.5(0.067 - x) = y (equation 1)
x + y = 0.268 (equation 2)
To solve these equations, we can use substitution or elimination method.
Let's start with substitution:
Rearrange equation 2 for y:
y = 0.268 - x
Substitute this into equation 1:
2.5(0.067 - x) = 0.268 - x
Simplify and solve for x:
0.1675 - 2.5x = 0.268 - x
-2.5x + x = 0.268 - 0.1675
-1.5x = 0.1005
x = 0.1005 / -1.5
x = -0.067 M
Since concentration cannot be negative, we discard this negative answer.
Therefore, the concentration of butane ([butane]) at equilibrium is 0.067 M.
Substitute this value into equation 2 to find the concentration of isobutane ([isobutane]):
0.067 + [y] = 0.268
[y] = 0.268 - 0.067
[y] = 0.201 M
Therefore, the concentrations at equilibrium are:
A) [butane] = 0.067 M
B) [isobutane] = 0.201 M