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Chapter 12: Problem 5

Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{I}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{IF}_{5}(g)\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)\) (d) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
Question: Write the equilibrium constant expressions for the following chemical reactions: a) I2(g) + 5 F2(g) ⇌ 2 IF5(g) b) CO(g) + 2 H2(g) ⇌ CH3OH(l) c) 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(l) + 2 SO2(g) d) SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(l) Answer: a) \(K = \frac{[\mathrm{IF}_{5}]^2}{[\mathrm{I}_{2}][\mathrm{F}_{2}]^5}\) b) \(K = \frac{[\mathrm{CH}_{3}\mathrm{OH(l)}]}{[\mathrm{CO(g)}][\mathrm{H}_{2}(g)]^2}\) c) \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2[\mathrm{SO}_{2}(g)]^2}{[\mathrm{H}_{2}\mathrm{S(g)}]^2[\mathrm{O}_{2}(g)]^3}\) d) \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2}{[\mathrm{H}_{2}(g)]^2}\)

Step by step solution

01

Identify the reactants and the products

The reactants in this reaction are I2(g) and F2(g), while the product is IF5(g).
02

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{IF}_{5}]^2}{[\mathrm{I}_{2}][\mathrm{F}_{2}]^5}\). #b) CO(g) + 2 H2(g) ⇌ CH3OH(l)#
03

Identify the reactants and the products

The reactants in this reaction are CO(g) and H2(g), while the product is CH3OH(l).
04

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{CH}_{3}\mathrm{OH(l)}]}{[\mathrm{CO(g)}][\mathrm{H}_{2}(g)]^2}\). Note that the concentration of the liquid CH3OH is considered as 1 in the reaction, because liquids and solids are considered to be in a separate phase that doesn't affect the equilibrium constant. #c) 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(l) + 2 SO2(g)#
05

Identify the reactants and the products

The reactants in this reaction are H2S(g) and O2(g), while the products are H2O(l) and SO2(g).
06

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2[\mathrm{SO}_{2}(g)]^2}{[\mathrm{H}_{2}\mathrm{S(g)}]^2[\mathrm{O}_{2}(g)]^3}\). Again, note that the concentration of the liquid H2O is considered as 1 in the reaction since liquids do not change the equilibrium constant. #d) SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(l)#
07

Identify the reactants and the products

The reactants in this reaction are SnO2(s) and H2(g), while the products are Sn(s) and H2O(l).
08

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2}{[\mathrm{H}_{2}(g)]^2}\). As before, the concentrations of the solid SnO2 and Sn, as well as the liquid H2O, are considered as 1 in the reaction.

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Most popular questions from this chapter

. Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ At a certain temperature, the equilibrium constant for the reaction is \(0.0255\). What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both reactants and products) is \(0.300 \mathrm{~atm} ?\)

Consider the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ When this system is at equilibrium at \(700^{\circ} \mathrm{C}\) in a \(2.0\) - \(\mathrm{L}\) container, \(0.10 \mathrm{~mol}\) \(\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}\), and \(0.40 \mathrm{~mol} \mathrm{C}\) are present. When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. Calculate \(K\) at \(700^{\circ} \mathrm{C}\) and again at \(600^{\circ} \mathrm{C}\).

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

At \(25^{\circ} \mathrm{C}, K=2.2 \times 10^{-3}\) for the reaction $$\mathrm{ICl}(g) \rightleftharpoons \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ Calculate \(K\) at \(25^{\circ} \mathrm{C}\) for (a) the decomposition of ICl into one mole of iodine and chlorine. (b) the formation of two moles of \(\operatorname{ICl}(g)\).

A compound, \(\mathrm{X}\), decomposes at \(131^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ If a flask initially contains \(\mathrm{X}, \mathrm{A}\), and \(\mathrm{C}\), all at partial pressures of \(0.250 \mathrm{~atm}\), in which direction will the reaction proceed?
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