Back to marketplace

Topic 11 - Equillibrium 2

No description yet. This curated deck is ready for daily study.

Total cards: 19•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (19)

#1
Front (Question)

What is Kp?

Back (Answer)

equilibrium constant in terms of pressure of gases

#2
Front (Question)

Kp=

Back (Answer)

p(products) _________ p(reactants) -only include gases

#3
Front (Question)

What is partial pressure?

Back (Answer)

the force each individual gas exerts on the walls of a container if it were alone

#4
Front (Question)

partial pressure =

Back (Answer)

mole fraction x total pressure

#5
Front (Question)

What is mole fraction?

Back (Answer)

the fraction of each gas moles of each gas out of total moles of gas

#6
Front (Question)

total pressure =

Back (Answer)

Σpartial pressures

#7
Front (Question)

What affects the value of the equilibrium constant?

Back (Answer)

temperature

#8
Front (Question)

What doesn’t affect the value of the equilibrium constant?

Back (Answer)

• changes in concentration • changes in pressure • presence of catalyst

#9
Front (Question)

What affect does an increase in temperature have on the value of Kc/Kp in an exothermic reaction?

Back (Answer)

K decreases

#10
Front (Question)

Why does Kc/Kp decrease for an exothermic reaction when there is an increase in temperature?

Back (Answer)

If T increases… ΔSsurr decreases so ΔStot decreases so K decreases

#11
Front (Question)

What affect does a decrease in temperature have on the value of Kc/Kp in an exothermic reaction?

Back (Answer)

K increases

#12
Front (Question)

Why does Kc/Kp increase for an exothermic reaction when there is a decrease in temperature?

Back (Answer)

If T decreases… ΔSsurr increases so ΔStot increases so K increases

#13
Front (Question)

What effect does an increase in temperature have on the value of Kc/Kp in an endothermic reaction?

Back (Answer)

K increases

#14
Front (Question)

Why does Kc/Kp increase for an endothermic reaction when there is an increase in temperature?

Back (Answer)

If T increases… ΔSsurr increases so ΔStot increases so K increases

#15
Front (Question)

What effect does a decrease in temperature have on the value of Kc/Kp in an endothermic reaction?

Back (Answer)

K decreases

#16
Front (Question)

Why does Kc/Kp decrease for an endothermic reaction when there is an decrease in temperature?

Back (Answer)

If T decreases… ΔSsurr decreases so ΔStot decreases so K decreases

#17
Front (Question)

What effect does an increase in temperature have on the position of equilibrium for an exothermic reaction?

Back (Answer)

eqm shifts left because… If T increases, K decreases so concentration term no longer equals new, smaller K value so p(products) must decrease and p(reactants) must increase to decrease the concentration term to do this equilibrium shifts left until the concentration term re-equals the value of K

#18
Front (Question)

What effect does an increase in temperature have on the position of equilibrium for an endothermic reaction?

Back (Answer)

equilibrium shifts right because… If T increases, K increases so the concentration term no longer equals the new, larger K value so p(reactants) must increase and p(products) must decrease to increase the concentration term to do this equilibrium shifts right until the concentration term re-equals K

#19
Front (Question)

What effect does an increase in pressure have on the position of equilibrium?

Back (Answer)

when P increases, all the partial pressures increase but by different amounts (larger powers = increase by greater amount) this means the concentration term no longer equals Kp -if there is a larger increase for the denominator (aka reactants), the denominator must decrease and the nominator must increase so that the concentration term increases and eqm shifts left until the concentration term re-equals Kp -if there is a larger increase for the nominator (aka products), the nominator must decrease and the denominator must increase so that the concentration term decreases and eqm shifts right until the concentration term re-equals Kp