\(PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}\) In the reaction above, an increase in pressure will...
In the reaction above, an increase in pressure will
- A) decelerate the reaction
- B) Increase the yield of PCl3
- C) increase the yield of PCl5
- D) accelerate the reaction
Correct Answer: C) increase the yield of PCl5
Explanation
In the given reaction, \(PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}\), we can see that one mole of \(PCl_5\) gas is decomposing into one mole of \(PCl_3\) gas and one mole of \(Cl_2\) gas. This means that the total number of moles of gas is increasing in this reaction.
According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in pressure, the system will adjust itself to counteract the change. In this case, an increase in pressure will cause the system to shift the equilibrium in the direction of the reaction that produces fewer moles of gas, in order to reduce the pressure.
Since the forward reaction (formation of \(PCl_3\) and \(Cl_2\)) produces more moles of gas, the reverse reaction (formation of \(PCl_5\)) will be favored when pressure is increased. Therefore, an increase in pressure will increase the yield of \(PCl_5\), as stated in Option C. It's essential to note that the reaction speed is not affected by the pressure change, so options A and D don't apply.