# Chapter 15: Applications of Eq. Constants ### Supplemental Instruction Iowa State University

Kelsey Chemistry 178 Verkade 09/14/2010 1.

Given the following information: 𝐻𝐹 (𝑎𝑞) → ← 𝐻 + (𝑎𝑞) 𝐻 2 𝐶 2 𝑂 4(𝑎𝑞) → ← 2𝐻 + (𝑎𝑞) + 𝐹 − (𝑎𝑞) 𝐾 𝐶 = 6.8 × 10 + 𝐶 2 𝑂 2− 4(𝑎𝑞) 𝐾 𝐶 −4 = 3.8 × 10 −6 determine the value of K C for the reaction: 2𝐻𝐹 (𝑎𝑞) + 𝐶 2 𝑂 2− 4(𝑎𝑞) → ← 2𝐹 − (𝑎𝑞) + 𝐻 2 𝐶 2 𝑂 4(𝑎𝑞) Kc = ((6.8 x 10^-4)^2)x(1/(3.8 x 10^-6) 2.

The density of Tin (Sn) is 7.287 g/cm 3 , what is the concentration (molarity, M)? 7.287 g/cm^3 = 7.287 g/ml (7.287 g/mL)*(1000 mL/1L)*(1 mol Sn/ 118.7 g Sn) = X M 3.

Write the equilibrium constant expression for K C for each of the following reactions: a.

𝐶𝑂 2(𝑔) + 𝐻 2(𝑔) → ← 𝐶𝑂 (𝑔) + 𝐻 2 𝑂 (𝑙) Kc=([CO]/([H2][CO2]} b.

𝑆𝑛𝑂 2(𝑠) + 2𝐶𝑂 (𝑔) → ← 𝑆𝑛 (𝑠) Kc=([CO2]^2)/([CO]^2) + 2𝐶𝑂 2(𝑔) 4.

A closed system initially containing 1.000 x 10 -3 M H 2 and 2.000 x 10 -3 M I 2 at 448 degrees C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 -3 M. Calculate K C at 448 degrees C for the reaction taking place, which is: 𝐻 2(𝑔) + 𝐼 2(𝑔) → ← 2𝐻𝐼 (𝑔) Make ICE table Kc=(Equilibrium Concentration of H2 * Equilibrium Concentration of I2)/(Equilibrium Concentration of HI squared)

### Supplemental Instruction

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5.

At 448 degrees C the equilibrium constant K C for the reaction 𝐻 2(𝑔) + 𝐼 2(𝑔) → ← 2𝐻𝐼 (𝑔) is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448 degrees C if we start with 2.0 x 10 -2 mol of HI, 1.0 x 10 -2 mol of H 2 , and 3.0 x 10 -2 mol of I 2 in a 2.00-L container. Reaction will proceed toward Products (Q

Consider the equilibrium 𝑁 2 𝑂 4(𝑔) → ← 2𝑁𝑂 2(𝑔) ∆𝐻° = 58.0 𝑘𝐽 In which direction will the equilibrium shift when a) N 2 O 4 is added, b) NO 2 is removed, c) the total pressure is increased by addition of N 2(g) , d) the volume is increased, e) the temperature is decreased? a) b) c) d) e) Right Right Left Right Right