Partial pressures and equilibrium question

In summary, the conversation discusses a reaction with a given equilibrium constant and partial pressures of the reactants. It then outlines a calculation to determine the equilibrium partial pressures of the gases involved. The attempted solution involves using a quadratic equation, but there are errors in the calculations. The correct answer should be 0.0546, 0.0077, and 0.0077 atm.
  • #1
veitch
7
0

Homework Statement



For the reaction 2HI (g) <-- --> H2 (g) + I2 (g) Kp = .02 at 445 degrees C
Parital pressures are .05, .01 and .01 respectively. Perform a calc to see which way it goes then figure out the equilibrium partial pressures of the gasses

The Attempt at a Solution



Qp = (.01)(.01)/(.05^2) = .0001/.0025 = .04

Since Qp>Kp reaction goes left

I .05 .01 .01
C +2x -x -x
E .05+2x .01-x .01-x

.02 = (.01-x)^2/(.05+2x)
0 = -x^2 + .06x + .0009

Quadratic equation yields -.0264 and -.0336... neither of which make sense as the final answers are given and they should be:
.0546, .0077 and .0077 atm
Which makes x = .0023...

That was my third time running through this problem and each time I got a different answer... no idea how I screwed this one up. it's only the first problem on an assignment due tomorrow... any help would be greatly appreciated.
 
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  • #2
veitch said:
.02 = (.01-x)^2/(.05+2x)

This is OK.

Quadratic equation yields -.0264 and -.0336

This is not. Just check your math.
 
  • #3
Borek said:
This is not. Just check your math.

You're right.. it should be 0= x^2 -.06x -.0009
Which has roots of 0.072 and -0.012 ...

But .05 + 2(.072) = 0.194 yet it should equal .0546 atm :(
 
  • #4
while (x != 0.0023) check_your_math;
 

FAQ: Partial pressures and equilibrium question

What is the meaning of partial pressure in chemistry?

Partial pressure in chemistry refers to the pressure exerted by an individual gas in a mixture of gases. It is proportional to the number of moles of the gas and the temperature, and is independent of the presence of other gases in the mixture.

How is partial pressure calculated?

Partial pressure can be calculated by multiplying the total pressure of a gas mixture by the mole fraction of the gas in the mixture. Alternatively, it can also be calculated by using the ideal gas law, where the partial pressure is equal to the product of the gas's molar concentration and the universal gas constant multiplied by the temperature.

What is the relationship between partial pressure and equilibrium?

In a gaseous reaction at equilibrium, the partial pressure of each gas is directly proportional to its equilibrium constant. This means that if the partial pressure of a gas increases, the equilibrium constant will also increase, and vice versa.

How does changing the partial pressure of a gas affect equilibrium?

Changing the partial pressure of a gas can affect the equilibrium of a reaction in two ways. First, if the partial pressure of a reactant is increased, the reaction will favor the forward reaction, shifting towards the products. Conversely, if the partial pressure of a product is increased, the reaction will favor the reverse reaction, shifting towards the reactants.

Can partial pressures ever be negative?

No, partial pressures cannot be negative as they represent the pressure exerted by a gas in a mixture. Negative pressure would imply that the gas is exerting a suction or pulling force, which is not possible.

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