Question

For which of the following reactions at equilibrium at constant temperature, doubling the volume will cause a shift to the right?

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
• B. $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
• C. $2CO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2C{O}_2\left(g\right)$
• D. $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
B $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

For the reactions (A) and (B) at equilibrium at constant temperature, doubling the volume will cause a shift to the right as these reactions proceed with increase in the number of moles of gaseous species. When volume is doubled, the pressure is reduced to half and the reaction will proceed in forward direction so that more products are formed and the effect of increased pressure is nullified.
The reaction (C) proceeds with decrease in the number of moles of gaseous species. hence, when the volume is doubled, the pressure is reduced to half and the reaction will shift in the backward direction.
For the reaction (C), the number of moles of gaseous reactants are equal to the number of moles of gaseous products. Hence, doubling the volume will have no effect on the equilibrium reaction.