Chapter 6: Problem 10
Under what pressure must an equimolar mixture of \(\mathrm{Cl}_{2}\) and \(\mathrm{PCl}_{3}\) be place at \(250^{\circ} \mathrm{C}\) in order to obtain \(75 \%\) conversion of \(\mathrm{PCl}_{3}\) into \(\mathrm{PCl}_{5}\) ? Given: \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{PCl}_{5}(\mathrm{~g}) ; K_{\mathrm{p}}=2 \mathrm{~atm}^{-1}\) (a) \(12 \mathrm{~atm}\) (b) \(6 \mathrm{~atm}\) (c) 15 atm (d) \(30 \mathrm{~atm}\)
Short Answer
Step by step solution
Understand the Chemical Equilibrium
Set Up the Equilibrium Expression
Determine the Initial Moles and Change
Express the Partial Pressures in Terms of Moles
Substitute the Partial Pressures into the Equilibrium Expression
Solve for the Total Pressure
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