1. Consider the following reaction: H2 (g) + Cl2 (g) ? 2 HCl (g) You have two fl

Question

1. Consider the following reaction: H2 (g) + Cl2 (g) ? 2 HCl (g) You have two flasks. Flask A IS at equilibrium and has partial pressures of PH2 = 0.958 atm, PCl2 = 0.877 atm and PHCl = 0.020 atm at 175 Kelvin. In Flask B, you have the following: PH2 = 0.621 atm, PCl2 = 0.621 atm and PHCl = 0.101 atm also at 175 Kelvin. a.) is reaction flask B at equilibrium? (explain how you know) b.) If not, which way will the reaction in flask B go to reach equilibrium and explain why. ( just telling me which direction it will go will not earn full points WHY must it go that direction to reach equilibrium

Explanation / Answer

A)

Flask A

Calculate the Ke value:

[0.02]^2 / ([0.958]*[0.877]) = 4.76 * 10^-4

Flask B

Calculate the Q value:

[0.101]^2 / ([0.621][0.621]) = 2.65*10^-2

Since the ratio in Flask B is not equal to that of flask A, it means that flask B is not in equilibrium. The Ke is the equilibrium constant for that reaction and any time that reaction is in equilibrium, the Ke value will be the same.

B)

The reaction in flask B will favor formation of the products (it will go to the left). We know this because the Q value calculated for flask B was higher that the Ke. That means that more of the products will be formed.

C)

Now,

Ke = 4.76*10^-4 = (0.101 - 2x)^2 / (0.621 + x)^2

Solve for x:

x=0.0432538

[HCl] = 0.0145

[H2]=[Cl2] = 0.664

H2 + Cl2 <-> 2HCl I 0.621 0.621 0.101 C x x -2*x E 0.621 + x 0.621+x 0.101 - 2x

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