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Chapter 13: Q48PE (page 755)

How can the pressure of water vapor are increased in the following equilibrium?

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\) \(\Delta H = 41kJ\)

Short Answer

Expert verified

The pressure of water vapor increased in the above-mentioned equilibrium by

(i) Increase in temperature

(ii) Increase in the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{l}})\).

Step by step solution

01

Understanding equilibrium

A physical change

\({\rm{\Delta H}} = 41{\rm{k}}.{\rm{J}}\)

Equilibrium can be established for a physical change as well as for a chemical reaction.

02

Increasing pressure of water vapor

Let us see how the pressure of water vapor can be increased.

(i) Since \({\rm{\Delta H}} > 0\)the reaction is endothermic, hence

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\)

By increasing the temperature (heat), the equilibrium of a physical change will move to the right and the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{g}})\) will increase; hence the pressure of water vapor will increase.

(ii) By increasing the amount of\({{\rm{H}}_2}{\rm{O}}({\rm{l}})\), the equilibrium of a physical change will move to the right, and the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{g}})\) will increase; hence the pressure of water vapor will increase.

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Most popular questions from this chapter

Question: A 1.00-L vessel at 400 掳C contains the following equilibrium concentrations: N2, 1.00M; H2, 0.50M; and NH3, 0.25M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1M?

Question: Consider the equilibrium

4NO2(g) + 6H2 O(g) 鈬� 4NH3(g) + 7O2(g)

(a) What is the expression for the equilibrium constant (Kc) of the reaction?

(b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?

(c) If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO2?

(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?

Question:At 1 atm and \(2{5^o}C,N{O_2}\) with an initial concentration of \(1.00M\;is\;3.3 \times 1{0^{ - 3}}\% \)decomposed into \(NO\;and\;{O_2}\). Calculate the value of the equilibrium constant for the reaction. \(2N{O_2}(g) \rightleftharpoons 2NO(g) + {O_2}(g)\)

What is the pressure of \(C{O_2}\)in a mixture at equilibrium that contains \(0.50atm\)\({H_2}\), \(2.0atm\)of \({H_2}O\), and \(1.0atm\)of \(CO\) at \(99{0^0}C\)?

\({H_2}(g) + C{O_2}(g) \rightleftharpoons {H_2}O(g) + CO(g)\)

\({K_P} = 1.6\,at\, 99{0^o}C\)

The following equation represents a reversible decomposition:

\({\mathbf{CaC}}{{\mathbf{O}}_3}{\text{ }}\left( {\mathbf{s}} \right) \rightleftharpoons {\mathbf{CaO}}\left( {\mathbf{s}} \right){\text{ }} + {\text{ }}{\mathbf{C}}{{\mathbf{O}}_2}{\text{ }}\left( {\mathbf{g}} \right):\)

Under what conditions will decomposition in a closed container proceed to completion so that no \({\bf{CaC}}{{\bf{O}}_3}\) remains?

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