# kc kp

Watch
Announcements
#1
Hi,
Sorry if I sould really stupid (I am lol), but can anyone kindly explain to me:

Why do Kc & Kp only change with temperature, and not with changes in pressure?

e.g - if we have 0.5 N2 + 1.5 H2 ⇌ NH3 (forward reaction exo)

- If I increase temp, posn eqm shifts left to oppose change & lower temp, which means [reactants] > [products] so kc/kp will decrease
- If I decrease temp, posn eqm shifts right to oppose change & increase temp, which means [products]> [reactants] so kc/kp increase

Why can't I apply the same logic to pressure changes?

Many thanks
0
10 months ago
#2
(Original post by not_XIX_forever)
Hi,
Sorry if I sould really stupid (I am lol), but can anyone kindly explain to me:

Why do Kc & Kp only change with temperature, and not with changes in pressure?

e.g - if we have 0.5 N2 + 1.5 H2 ⇌ NH3 (forward reaction exo)

- If I increase temp, posn eqm shifts left to oppose change & lower temp, which means [reactants] > [products] so kc/kp will decrease
- If I decrease temp, posn eqm shifts right to oppose change & increase temp, which means [products]> [reactants] so kc/kp increase

Why can't I apply the same logic to pressure changes?

Many thanks
Hey,
If you increase concentration on the rhs for example equilibrium will shift to the lhs to decrease this, meaning although the amount of reactants and products have changed the ratio stays constant. Therefore, no net change, so the kc and kp values do not change- it's the same for pressure as well.
hope this helps
0
10 months ago
#3
Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, k, as shown in the Arrhenius equation).

The equilibrium constant, Kc is the ratio of the rate constants, so only variables that affect the rate constants can affect Kc. Pressure doesn't show in any of these relationships.

What is the definition of equilibrium? Forward Rate == Backward Rate (essentially).

What influences the rate of reaction? k? Concentration (e.g. rate = k[A][B]), how many collide with the correct orientation, temperature, and activation energy (see Arrhenius equation).

Kc is a ratio of [Product]/[Reacant]. You don't see pressure appear in any of these equations (rate law or Arrhenius).

Pressure may change the rate of reaction, but not the rate constant / equilibrium constant.

This can be difficult to get your head around initially. You might find it easier to look at the equations and figure it out mathematically - this helped me.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (207)
57.66%
I don't have everything I need (152)
42.34%