How Do Different Stresses Affect a Chemical Equilibrium Reaction?

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lapo3399
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1.
Given the equilibrium reaction:
N2(g) + O2(g) <---> 2NO(g) deltaH=180kJ mol-1
State the effect of each of the following potential stresses on the system:
A) Temperature is decreased
B) The pressure of the container is decreased, keeping volume constant
C) A catalyst is added
D) NO is removed as it forms
E) Argon gas is added to the container

2.
A) A shift to the left will occur, considering the enthalpy of reaction.
B) Not sure about this one, as the only way to accomplish this is to remove particles from the container. My best guess is that, assuming an equal quantity of reactants vs. products is removed, there will be no shift as the ratio of moles of reactant to moles of product is 1:1.
C) No change, as the catalyst will catalyze both the forward and reverse reactions equally.
D) Shift to the right, as product collisions decrease as NO is removed.
E) No change, as the argon will increase the total pressure, but will not react in collisions.
 
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Gokul43201 said:
For instance, addition of argon (at constant volume) to the reaction N2 + 3H2 <--> 2NH3, will produce a shift to the right.

Why is this? Argon is inert and the partial pressures of the reactants and product will remain unchanged.
 
Plastic Photon said:
Why is this? Argon is inert and the partial pressures of the reactants and product will remain unchanged.

Well you are talking about stressing the equilibrium point. if the equation increases pressure if it goes to the right, and you increase the pressure the equation will go towards relieving this pressure and will go to the left. and I'm sure about this answer.
And if the equation increases temperature as it goes to the right, then a decrease in temperature will goes it to increase the temperature even more. cause it has to maintain its stability.
 
eaboujaoudeh said:
Well you are talking about stressing the equilibrium point. if the equation increases pressure if it goes to the right, and you increase the pressure the equation will go towards relieving this pressure and will go to the left. and I'm sure about this answer.
And if the equation increases temperature as it goes to the right, then a decrease in temperature will goes it to increase the temperature even more. cause it has to maintain its stability.

You can see for youself why this cannot possibly be true if you substitute the concentration of the argon into the equilibrium expression. The same term will appear in the numerator and denominator of the equilibrium expression. Cancelling gives you an equilbrium expression that is only dependent upon the concentrations of products and reactants. Argon is neither reactant nor product.
 
ok then. what happens if i increase the pressure in the container without adding or changing anything else?
 
N2 + 3H2 <--> 2NH3 i meant for this equation sorry. if i increase pressure here i would get the equation shifting towards the right to relieve the pressure right? how do u increase pressure? well u induce a nonreactive gas! this gas increases the overall pressure in the container thus it stresses the equation and causes it to move towards the pressure releasing side.
 
thus my quote was right ! i was quoting plastic photon who said that adding a nonreactive gas to an equation won't alter its' equilibrium pt.my answer is that it will !
 
An addition of inert gas changes the total pressure, however an addition of reactive gas changes both the total pressure and the partial pressure of the reactants and products.
 
chemisttree said:
Absolutely.

Sorry, I should have replied "Absolutely not."


Will increasing the pressure of a vessel with an inert gas change the concentration of either the reactants or products?
 
woops..i guess you're right. adding an inert gas will have no effect on the concentration of products and reactants..ssorryy