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Chapter 13: Q33 E (page 752)

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

Expert verified

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

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.

Question: Butane exists as two isomers, n鈭抌utane and isobutane.

\({K_P} = 2.5\;at\;2{5^o}C\)

What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?

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) + H2翱(濒)鉄禖6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 脳 10鈭�11/s at 27 掳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鈥攖he 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 脳 105 at 27 掳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.

Analysis of the gases in a sealed reaction vessel containing \(N{H_3}\), \({N_2}\), and \({H_2}\) at equilibrium at \(40{0^0}C\) established the concentration of \({N_2}\) to be \(1.2M\) and the concentration of \({H_2}\) to be \(0.24M\).

\({N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)\)

\({K_c} = 0.50\,at\,40{0^o}C\)

Calculate the equilibrium molar concentration of \(N{H_3}\).

Round the following to the indicated number of significant figures:

(a) 0.424 (to two significant figures)

(b) 0.0038661 (to three significant figures)

(c) 421.25 (to four significant figures)

(d) 28,683.5 (to five significant figures)
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