What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?

Short Answer

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Changes in enthalpy may be used.

Step by step solution

01

Step 1: What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant.

When there is an exothermic reaction, the produced heat is considered as a product.

But when there is an endothermic reaction, the heat is considered to be a reactant.

Hence, changes in enthalpy may be used. An exothermic reaction may be shifted to reactants due to additional heat but there will be a shift to the product side when it comes to endothermic reactions.

However, cooling down the endothermic reaction will make it shift towards the reactants instead.

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Most popular questions from this chapter

Which of the systems described in Exercise 13.15 give homogeneous equilibria? Which give heterogeneous equilibria?

(a) \(C{H_4}(g) + C{l_2}\rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b)\({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

(c)\(2S{O_2}(\;g) + {O_2}(\;g)\rightleftharpoons 2S{O_3}(\;g)\)

(d)\(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f)\(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h) \(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation \({\rm{KI}}(aq) + {{\rm{I}}_2}(aq) \rightleftharpoons {\rm{K}}{{\rm{I}}_3}(aq)\) give the same expression for the reaction quotient. \({\rm{K}}{{\rm{I}}_3}\)is composed of the ions \({{\rm{K}}^ + }\) and \({{\rm{I}}_3}^ - .\)

Question: In a 3.0-L vessel, the following equilibrium partial pressures are measured: \({{\rm{N}}_2}\),190 torr;\({{\rm{H}}_2}\), 317 torr;\({\rm{N}}{{\rm{H}}_3}\)\(1.00 \times {10^3}\)torr.

  1. How will the partial pressures of\({{\rm{H}}_2},{{\rm{N}}_2}\)and \({\rm{N}}{{\rm{H}}_3}\)change if \({{\rm{H}}_2}\) is removed from the system? Will they increase, decrease, or remain the same?
  2. Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide (\({\rm{S}}{{\rm{O}}_3}\)), from sulfur dioxide (\({\rm{S}}{{\rm{O}}_2}\)), and oxygen (\({{\rm{O}}_2}\)), shown here.

\(2{\text{S}}{{\text{O}}_2}(g) + {{\text{O}}_2}(g) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}(g)\)

At high temperatures, the rate of formation of \({\rm{S}}{{\rm{O}}_3}\)is higher, but the equilibrium amount (concentration or partial pressure) of \({\rm{S}}{{\rm{O}}_3}\) is lower than it would be at lower temperatures.

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

(b) Is the reaction endothermic or exothermic?

Question: The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

C12 H22 O11(aq) + H2O(l)⟶C6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

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