Equilibrium 1
At a particular temperature, a 3.0-L flask contains 3.50 mol HI, 4.10 mol H2, and 0.300 mol I2 in equilibrium. Calculate Kc at this temperature.
H2 (g) + I2 (g) ⇌ 2 HI (g)
Convert to Molarity:
H2 (g)= 1.37 M, I2 (g)= 0.100 M, HI= 1.17 (g)
Kc= (1.17)2/ ((1.37)(0.100))= 10.0
How is the position of equilibrium changed by the addition of a catalyst? How is the equilibrium constant changed by the addition of a catalyst?
A catalyst does not affect either the position of equilibrium or the equilibrium constant for a given reaction. A catalyst speeds the forward and reverse reactions equally by lowering their activation energy. The only thing that changes an equilibrium constant is a change in temperature.
Classify the following as strong acids (SA), weak acids (WA), strong bases (SB), or weak bases (WB)
HI
KOH
HF
NH3
(CH3)3NH+
SA
SB
WA
WB
WA
For this equilibrium, what change to pressure will favor formation of products? What change to temperature will favor formation of products?
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) Δ H= -92 kJ
Decrease in temperature, because heat can be viewed as a product in this exothermic reaction:
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) + 92 kJ
Decreasing the temperature will subsequently cause a decrease in the rate of the reverse reaction.
Increase in pressure, because Δn= -2 for this reaction. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favoring the reaction which produces the fewer molecules.
A (g) + B (g) ⇌ C (g) Δ H= + 53.9 kJ
For the above reaction, how will the equilibrium be affected by:
a) The addition of an inert gas, such as He,
b) An increase in temperature,
c) The volume of the container that holds the equilibrium mixture being decreased.
a) No effect. An inert gas will not cause any reaction, and will not change the partial pressures of the products and reactants.
b) A shift toward the products will occur. The reaction is endothermic; therefore, addition of heat will cause the forward reaction to increase.
c) A shift toward the products will occur. A decrease in volume will increase the pressure, causing the reaction to attempt to regain the lower pressure. (see Le Chatelier's Principle) There are less gas molecules on the product's side, so the forward reaction will be favored.
A solution has a hydronium ion concentration of 2.0 × 10−6 M. What is the concentration of hydroxide ion at 25°C?
2 H2O(l) ⇌ H3O+ (aq) + OH−(aq)
KW = [H3O+][OH-]
[OH-] = Kw/[H3O+] = 1.00 x 10-14/2.0 × 10−6
[OH-] = 5.0 × 10−9 M
I2(g) + Br2 (g) ⇌ 2 IBr (g)
2.00 M of I2 and 2.00 M of Br2 are initially present in a reaction mixture. Calculate the equilibrium concentration of IBr at 425 K. Kc= 100.0
R I2(g) + Br2 (g) ⇌ 2 IBr (g)
I 2.00 2.00 0
C -x -x +2x
E 2-x 2-x +2x
Kc= 100= (2x)2/ ((2-x)(2-x)) Take the square root of both sides of the equation:
10= 2x / (2-x) ; x= 1.67 M
[IBr] at equilibrium= 2(1.67)= 3.34 M
The partial pressure of SO2, O2, and SO3 are 0.286 atm, 0.414 atm, and 1.28 atm, respectively. Kp= 86.1. Determine if the reaction is at equilibrium. If not, which direction will the reaction shift?
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
Q= (1.28)2/ ((0.414)(0.286)2) = 48.8
Q<Kp, so reaction is not at equilibrium. It will shift toward the products side (forward reaction will proceed more quickly than the reverse reaction until equilibrium is attained).