Question:
<em>For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?</em>
Answer:
<em>The reaction will proceed towards the liquid phase. Heat is on the reactant side of the equation. Lowering temperature will shift equilibrium left, creating more liquid water. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.</em>
<em>If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium, that will be the back reaction because the forward reaction is exothermic.</em>
Hope this helps, have a good day. c;
1a 1
b 4
c 5
2 (5.44x10^2)(2.5x10^-3)(7.9x10^-3) = 1.1x10^-6
3 750/1000 = 0.750
4 50/0.5 = 100 mL
5 He came up with the modern ideals for chemsitty
Ultraviolet light is the correct answer
<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
Molal or molar? There is a difference