For each of the questions, four choices have been provided. Select the correct alternative. Equilibrium position of which of the following reactions is not affected by change in pressure? (a) \(\mathrm{I}_{2(\mathrm{~s})}+5 \mathrm{~F}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{IF}_{5(\mathrm{~g})}\) (b) \(\mathrm{FeO}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})} \rightarrow \mathrm{Fe}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})}\) (c) \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{CuO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}\)

Step by step solution

01

Examine reaction (a)

Compare the total moles of gaseous reactants and products in the equation: \(I_2(s) + 5F_2(g) \rightarrow 2IF_5(g)\) On the left side, we have 5 moles of gaseous \(F_2\). On the right side, we have 2 moles of gaseous \(IF_5\). The number of moles of gaseous species is different, thus pressure change will affect this reaction's equilibrium position.

02

Examine reaction (b)

Compare the total moles of gaseous reactants and products in the equation: \(FeO(s) + CO(g) \rightarrow Fe(s) + CO(g)\) On the left side, we have 1 mole of gaseous \(CO\). On the right side, we also have 1 mole of gaseous \(CO\). The number of moles of gaseous species is the same, thus pressure change won't affect this reaction's equilibrium position. It appears we have found the correct answer, but for completeness, let's examine the remaining reactions.

03

Examine reaction (c)

Compare the total moles of gaseous reactants and products in the equation: \(2Cu(NO_3)_2(s) \rightarrow 2CuO(s) + 4NO_2(g) + O_2(g)\) Reactants have no gaseous component. On the right side, we have 4 moles of gaseous \(NO_2\) and 1 mole of gaseous \(O_2\). The number of moles of gaseous species is different on each side, thus pressure change will affect this reaction's equilibrium position.

04

Examine reaction (d)

Compare the total moles of gaseous reactants and products in the equation: \(N_2O_4(g) \rightarrow 2NO_2(g)\) On the left side, we have 1 mole of gaseous \(N_2O_4\). On the right side, we have 2 moles of gaseous \(NO_2\). The number of moles of gaseous species is different, thus pressure change will affect this reaction's equilibrium position. Thus, the correct alternative is (b), where the equilibrium position of the reaction is not affected by changes in pressure.

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