Question

4. Using the information provided under the periodic table, calculate the free energy of reaction at 25.0 °C, assuming that a 1.00 L flask contains 0.125 moles NH4CI(s), 0.250 mols NH3(g), and 0.075 mols HCl(g) NH,Cl(s) 근 NH,(g) + HCl(g) a. 86 kJ/mol b. 91 kJ/mol c. -9.8 kJ/mol d. -4.6 kJ/mol e. 81 kJ/mol 5. What is the pH of 0.43 grams of strontium hydroxide (MM- 121.6 g/mol) dissolved in 10.0 liters of water? a. 3.15 b. 7.00 c. 10.85 d. 3.45 e. 10.55 6. Which of the following conditions will increase the production of COCl2(g) in the following reaction? CO(g) + Cl2(g) 근 COCMg) AH112 kJ/mol a. increase the temperature b. decrease the concentration of CO(g) c. increase the volume d. all of the above e. none of the above

Answer 1

Correct options: 4.(b) 5.(c) 6.(e)

​​​​​​EXPLAINATIONS:

6. The reaction is exothermic. So

NH_4 Cl (s) doubleheadarrow NH_3 (g) + HCl (g) delta G degree_r times n = delta G degree_product - delta G_reactant = delta G degree NH_3 (g) + delta G degree HCl (g) - delta G degree NH_4 Cl (3) = (negative 16.45 - 95.30 + 202.87) kJ/mol - 91.12 kJ/mol = 91 kJ/mol Hence the correct option is 6. Moles of strontium hydroxide = 0.43 g/121.6 g/mol = 0.003536 mol Molarity of strontium hydroxide = 0.003536 mol/10.0 L Dissociation of strontium hydroxide: Sr (OH)_2 (aq) rightarrow Sr^2 + (aq) + 2 OH^- (aq) [OH^-] = 2 times [Sr (OH)_2] = 2 times 0.0003536 M = 0.0007072 M POH = negative log [OH^-] = negative log(0.0007022) = 3.15 pH = 14 - pOH = negative log (0.0007022) = 3.15 pH = 14 - pOH = 14 - 3.15 = 10.85

• Increasing the temperature will shift the equilbrium towards reactant.
• CO is reactant. So, decreasing the concentration of CO will shift equilibrium towards reactant.
• Moles of gaseous reactant is greater than the moles of gaseous products. So, increasing the volume will shift the equilibrium towards reactant side.

Thus, none of the factors given in options a, b and c will favor the reaction forward. So, these factors will not increase the production of the product, COCl₂. Hence, the correct option is (e).