Question on chemical equilibrium. Law of mass action.


First of all, Ye shall not do my homework here! I do not quite understand here just a job and would appreciate a APPROACH.

The task is: The equilibrium constant of the conversion reaction at 500 ° C has the value Kc =. 4 What is the concentration of H2O in a GG mixture with c (CO2) = 11mm, c (H2) = 3 mM and c (CO) = 41mm?

When equation I've written: CO + H2O → CO2 + H2 (with double arrow). I hang now firmly in calculating the metabolic rate. How do I have the 11mm and 3 mM now use to allow the left side of the reaction to subtract?

Have just no insight. As I said: One approach would hand me.

Thank you in advance:)

The best answer

If I have your statement correctly understood, then

CO + H₂O ⇌ CO₂ + H₂

K = c (H₂) · c (CO₂) / (c (CO) · c (H₂O)) = 4

What is the concentration of H2O in a GG mixture with c (CO2) = 11mm, c (H2) = 3 mM and c (CO) = 41mm?

My feeling can mean two things. Either it is meant "in equilibrium is measured c (CO₂) = 11 mmol / l etc, how big is c (H₂O)?". Then you simply sets the four numbers (three concentrations and K) in the above equation and solving for c (H₂O) on.

Or is it meant "We pump 11 mmol / l CO₂ etc. into the reaction vessel and wait until the equilibrium is reached. What are then the concentrations of all 4 components? ". Moreover, you call your part, responsive, and x is to c (CO₂) = c₀ (CO₂) -x, H₂ well, c (CO) = c₀ (CO) + x, c (H₂O) = x. The purely into the equation, solving for x and calculate the four concentrations.

Hi, the MWG says yes "product of the concentrations of all products divided by the product of the concentrations of all reactants"

K = ([CO2] x [H2]) / ([CO] x [H2O]) = 4

But what mM? millimoles? Insert the concentrations Definitely and solve for [H2O]

lG, Chris

Adjusting the MWG on, releasing a known concentrations and ask for the unknown to.

Date: 2018-12-25 Views: 3

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