Question

At 700 $K,K_p=0.140$ for the reaction $CI{F}_3\left(g\right)\rightleftharpoons ClF\left(g\right)+F_2\left(g\right)$. Calculate the equilibrium partial pressure of $Cl{F}_3$, $ClF$, and $F_2$ if only $Cl{F}_3$ is present initially, at a partial pressure of 1.47 atm.

• A. $P_{ClF}=P_{F_2}=0.389atm,P_{Cl{F}_3}=1.08atm$
• B. $P_{ClF}=0.45atm,P_{F_2}=0.276atm,P_{Cl{F}_3}=2.30atm$
• C. $P_{ClF}=0.50atm,P_{F_2}=0.389atm,P_{Cl{F}_3}=1.67atm$
• D. $P_{ClF}=P_{F_2}=0.678atm,P_{Cl{F}_3}=3.04atm$

Answer 1

Answer: (A)

$P_{ClF}=P_{F_2}=0.389atm,P_{Cl{F}_3}=1.08atm$

	$Cl{F}_3$	$ClF$	$F_2$
Initial pressure (atm)	$1.47$	$0$	$0$
Final pressure (atm)	$1.47-x$	$x$	$x$

$Cl{F}_3\to ClF+F_2$

The expresion for equilibrium constant is $K_P=\frac{\left[ClF\right]\left[F_2\right]}{\left[Cl{F}_3\right]}$

$0.140=\frac{x\times x}{1.47-x}$

$x^2+0.140x-0.2058=0$

$\left[ClF\right]=\left[F_2\right]=x=0.389$ atm

$\left[Cl{F}_3\right]=1.47-x=1.47-0.389=1.08$ atm