University:
Jackson College
Course:
CEM-141 | General Chemistry I
Academic year:

2024
Views:

156

Pages:

19
Author:

fansiden4dut

EQUILIBRIUM CONSTANT EXPRESSIONS AND CALCULATIONS Special K traits: 1. K is a unitless constant 2. K is generated from equilibrium concentrations in MOLARITY units. If other units are used, a special subscript is applied to K to note that change. eg. if kPa units are used, the value would be Kp 3. K is a fixed constant for a given equilibrium system at a FIXED TEMP. Only a change in temperature can change K. Recall that entropy is favoured at high temp and enthalpy is favoured at low temp. Changes in Conc, Pressure, Volume, etc. may alter the concentration of each individual species, but this will not change K. K RTP + K STP (not equal!) 4. Species in the solid or liquid states are included in K expressions only if the equilibrium system is HOMOGENEOUS. In heterogeneous systems, they are excluded because their concentrations are invariable at a fixed temperature.< (i.e. density is constant at a fixed temperature) SEE GIVEN EXAMPLES. #4. examples "Keq" expressions Write the K expressions for the following. 5. H2 (g) + I2 (g) ⇌ 2HI (g) I 4.00 M + 2.00 M R -x -x +2x E 4-x 2-x 2x If Keq = 5.00 x 10^1 at 110°C, find equilibrium concentrations of hydrogen and iodine gas. Keq = 50 = (2x)^2 / (4-x)(2-x) LIMIT: 0 < x < 2 x = 1.869 [H2] = 2.13 M [I2] = 0.13 M 6. At a certain temperature, 0.040 M of phosphorus pentachloride was placed in a flask and allowed to come to equilibrium in a closed system. The concentration of each of the products at equilibrium was 0.025 M. Calculate Keq. PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) I 0.040 M R -0.025 +0.025 +0.025 E 0.015 0.025 M 0.025 M Keq = [PCl3][Cl2] / [PCl5] = (0.025)^2 / 0.015 = 4.2 x 10^-2 7. The same reaction was carried out at the same temperature (as in question #6), but a different starting concentration of reactants was used. Calculate [PCl5]i and [PCl5]eq. PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) I x R -0.10 +0.10 +0.10 E x-0.10 0.10 M 0.10 M SET LIMITS: x > 0.10 4.2 x 10^-2 = (0.10)^2 / (x-0.10) [PCl5]i = x = 0.34 M [PCl5]eq = 0.24 M 8. The reaction: 2XY(g) ⇌ X2(g) + Y2(g) has a Keq = 35 at 25°C. 3.0 moles of XY are injected into a 1.0 L container at 25°C, find the equilibrium [X2] and [Y2]. 2XY(g) ⇌ X2(g) + Y2(g) I 3.0 moles R -2x +x +x E 3.0 - 2x x x Keq = [X2][Y2] / [XY]^2 35 = x^2 / (3-2x)^2 5.916(3-2x) = x x = 1.383 M [X2]eq = [Y2]eq = 1.4 M 9. Given the equilibrium equation: 3A(g) + B(g) ⇌ 2C(g) If 2.50 moles of A and 0.500 moles of B are added to a 2.00 L container, an equilibrium is established in which the [C] is found to be 0.250 M. a) Find [A] and [B] at equilibrium. 3A(g) + B(g) ⇌ 2C(g) I 2.50moles 0.500 moles 2.00L 2.00L =1.25M 0.250M R -0.375 -0.125 +0.250 E 0.875 0.125 0.250 M 2 dec 3 dec 3 dec places places places Keq = [C]^2 / [A]^3[B] = (0.250)^2 / (0.875)^3(0.125) = 0.73 b) Calculate the value of the equilibrium constant Keq. K = 0.75 if you round it off to 2SF (if you use (0.875)^3)