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Kp multiple choice help

Can someone please help on how I can answer this Q: https://imgur.com/HIxCpMs

First off I know the reaction is exothermic due to a negative enthalpy value.

From directly looking at the given equation, everything seems to be in a 1:1 ratio as there's no coefficient numbers in front of any species. Therefore I won't be dealing with any raised powers when it comes to writing the Kp expression.
_______________________________________________________
My Kp expression is:

(P_WXY)(P_Y)/(P_W)(P_X)(P_Y^2)

I know that partial pressure is...

Mole fraction of gas * total pressure of mixture

where "mole fraction of gas" is number of moles of gas over total number of moles of gas in mixture.
______________________________________________________________

Looking at my options it definitely cannot be C.

If temperature increase, equilibrium will tend to endothermic or backward direction, meaning more moles of reactants. So partial pressures of W, X and Y2 increase leading to Kp reduce.

Maybe B, as there are more gaseous moles on LHS than RHS, so attempting to increase pressure means position of equilibrium favours side with less gas moles which is product side. This means partial pressures of both WXY and Y2 rise so Kp also goes up?

I have no clue with what happens with A and D?

Any help would be nice!

Reply 1

eznot

Kp values and Kc values ARE ONLY affected by temp.

change in pressure only affects equil position

Original post by Yatayyat

Reply 2

anonymoussse

the only thing which changes the value of kp and kc is temperature. so A

Reply 3

Yatayyat

Original post by Ezooner

Kp values and Kc values ARE ONLY affected by temp.

change in pressure only affects equil position

Ok so that eliminates D and is false, since temperature is clearly dependent of whether Kp increases or decrease.

I get that pressure must change equil position, because we have an equation where there is an imbalance of gaseous moles on either side. But does't a change in equil position affect Kp value then?

Reply 4

Yatayyat

Original post by ggxsywes

the only thing which changes the value of kp and kc is temperature. so A

But if lets say pressure was lower, then obviously equilibrium position would shift to favour side with more gas moles to oppose the reduction in pressure. Wouldn't that mean that partial pressure values of reactants increase and would be bigger than partial pressure values of product.

And since Kp is pp(products)/pp(reactants), then Kp would decrease. I just can't see how pressure is independent of not affecting Kp??

Reply 5

eznot

No, it doesn't I'm not sure I can explain why I've just learnt to accept the only temperature effects Kp

Original post by Yatayyat

Reply 6

Yatayyat

Original post by Ezooner

No, it doesn't I'm not sure I can explain why I've just learnt to accept the only temperature effects Kp

Ok well thank you anyway for the help. I'm just a bit baffled as to why a change in equilibrium position won't affect Kp. Surely it has to as it leads to change in pressure of both reactants and products.

Reply 7

anonymoussse

pressure does affect the kp value but the kp value changes TEMPORARILY. so let's say u change pressure of the system, the kp value changes too. but u want to reestablish the old kp value so the products/reactants shift in value to do so.

Original post by Yatayyat

But if lets say pressure was lower, then obvio[post="84001192"]save[/post]usly equilibrium position would shift to favour side with more gas moles to oppose the reduction in pressure. Wouldn't that mean that partial pressure values of reactants increase and would be bigger than partial pressure values of product.

And since Kp is pp(products)/pp(reactants), then Kp would decrease. I just can't see how pressure is independent of not affecting Kp??

(edited 4 years ago)

Reply 8

Yatayyat

Original post by ggxsywes

I see, so in a way Kp always want to return to its fixed value for when pressure alters. Only factor to permanently change the Kp value is altering temperature. I need to remember this. Thanks :smile:

Reply 9

anonymoussse

the reason why the kp value is constant is because if u increase the products for example u need to also simultaneously decrease the reactants by the same amount. this keeps kp value constant. https://www.chemguide.co.uk/physical/equilibria/change.html

i'm actually having trouble understanding how the kp value is affected by the pressure change. not when ure shifting the equilibrium and such....i mean in the beginning when the kp value temporarily changes. how does it change?!

a + b ----> 2c + 3d (lets pretend this is an equilibrium arrow and not a full arrow)
lets say u increase pressure. kp value changes temporarily. the products have more moles. so they are more affected. so in order to decrease pressure, the products decrease in amount and the reactants increase in amount. kp is now kept constant.

but i dont get how the value of kp was ever changed to begin with
@charco and @Pigster please help (OHHH i think it's because the kp value just signifies the relative partial pressures of the species in the reactants and the species in the products. thus, increasing pressure means the partial pressures of the numerator (products) INCREASES more than the denominator (reactants) because the products are higher in moles. therefore, the kp value increases. in order to reshift the kp value and the restore the partial pressures of the products, u can do so by decreasing the moles of the products because essentially, partial pressure=mole fraction* total pressure. therefore although u cant change the total pressure, u can still effectively restore the partial pressures by decreasing the molar fraction of the products. equilibrium shifts to the reactants, i.e, the reactants increase in amount and the kp value is restored! HOORAY!)

Original post by Yatayyat

(edited 4 years ago)

Reply 10

Pigster

You're best off avoiding this thinking/language.

Kp doesn't change.

What can change temporarily is the ratio of reactants to products. When the ratio does not equal Kp, the reaction is not at equilibrium. When the shift happens, it is to restore the ration to the value of Kp.

Original post by ggxsywes

therefore, the kp value increases.