Respuesta :
Answer:
See explanation below
Explanation:
To do this, let's write the reaction again:
H₂ + Br₂ <------> 2HBr Kc = 50.8
we are given the pressures, but we cannot use the pressure and the Kc values together, we need to convert Kc to Kp, and we do this with the following:
Kp = Kc(RT)^Δn
first, the value of Δn is calculated substracting the coefficients of the reaction:
Δn = 2 - (1+1) = 0
Kp = 50.8 * (0.082 * 298)⁰
Kp = 50.8 * 1 = 50.8
So Kc = Kp in this case. Now, let's draw an ICE chart, so we can know the equilibrium pressure of all compounds:
H₂ + Br₂ <------> 2HBr Kp = 50.8
i) 9.28 9.28 0
c) -x -x +2x
e) 9.28-x 9.28-x 2x
The equilibrium reaction will be:
Kp = PpHBr² / PpH₂ * PpBr₂
Replacing we have:
50.8 = (2x)² / (9.28-x)²
50.8 = 4x² / (86.11 - 18.56x + x²)
50.8(86.11 - 18.56x + x²) = 4x²
4374.39 - 942.85x + 50.8x² = 4x²
46.8x² - 942.85x + 4374.39 = 0
From here, we can use the quadratic equation general formula to get x, but we have some high numbers, so, let's reduce the numbers a little more, dividing by 46.8 the whole equation:
x² - 20.15x + 93.47 = 0 a = 1; b = -20.15; c = 93.47
The general equation is:
x = -b ±√b² - 4ac / 2a
Replacing the values we have:
x = 20.15 ±√(20.15)² - 4 * 1 * 93.47 / 2
x = 20.15 ±√32.14 / 2
x = 20.15 ± 5.67 / 2
x1 = 20.15 + 5.67 / 2 = 12.91
x2 = 20.15 - 5.67 / 2 = 7.24
As the innitial pressures are 9.28, the first value cannot be the value of the equilibrium pressure at the end, so, the second value of x, will be the correct one, therefore the partial pressures of the gases are:
[HBr] = 2x = 2 * 7.24 = 14.48 atm
[H₂] = [Br₂] = 7.24 atm
The equilibrium pressure of Br2 will be 7.24 atm.
What is equilibrium pressure?
The equilibrium pressure can be calculated by minus the x from initial pressure.
Given,
The reaction
[tex]\rm H_2 + Br_2 <---> 2HBr\\\\ Kc = 50.8[/tex]
We have to convert the kc to kp, because we can not use the kc value and pressure together.
[tex]Kp = Kc(RT)^\Delta^n[/tex]
Δn = 2 - (1+1) = 0
[tex]Kp = 50.8 \times (0.082 \times298)^\circ\\\\Kp = 50.8 \times 1 = 50.8[/tex]
Now Kc = Kp
Kp = 50.8
First drawing the ICE chart
i) 9.28 9.28 0
c) -x -x +2x
e) 9.28-x 9.28-x 2x
The equilibrium reaction will be:
[tex]Kp = \dfrac{PpHBr_2}{PpH_2\times PpBr_2} \\\\50.8 = \dfrac{(2x)^2}{ (9.28-x)^2} \\\\ 50.8 = \dfrac{4x^2}{ (86.11 - 18.56x + x^2)}\\\\50.8(86.11 - 18.56x + x^2) = 4x^2\\\\4374.39 - 942.85x + 50.8x^2 = 4x^2\\\\46.8x^2 - 942.85x + 4374.39 = 0[/tex]
[tex]x^2 - 20.15x + 93.47 = 0\\\\ a = 1; b = -20.15; c = 93.47[/tex]
Now, we can use the quadratic equation to get x
The equation is
[tex]x = -b+- \sqrt{b^2}- \dfrac{ 4ac }{2a}[/tex]
Putting the values
[tex]x = 20.15+- \sqrt{(20.15)^2}- \dfrac{ 4\times1 \times 93.47}{2}\\\\x = 20.15+- \dfrac{ \sqrt{(32.14)^2}}{2}\\\\x1 = 20.15 +\dfrac{5.67}{ 2} = 12.91\\\\ x2 = 20.15 - \dfrac{5.67}{ 2} = 7.24[/tex]
The value of x2 is taken
[tex][HBr] = 2x = 2 \times 7.24 = 14.48\; atm\\[/tex]
[tex][H_2] = [Br_2] = 7.24\; atm[/tex]
Thus, the equilibrium pressure is 7.24 atm.
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