In the reaction below, 9.28 atm each of H2 and Br2 were placed into a 1.00 L flask and allowed to react: H2(g) + Br2(g) <=> 2 HBr(g) Given that Kc = 50.8, calculate the equilibrium pressure of Br2 to 2 decimal places.

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.