Question

At a certain temperature, the reaction $$ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g) $$ gives a \(50.0 \%\) yield of \(\mathrm{XeF}_{4},\) starting with \(\mathrm{Xe}\left(P_{\mathrm{Xe}}=0.20 \mathrm{~atm}\right)\) and \(\mathrm{F}_{2}\left(P_{\mathrm{F}_{2}}=0.40 \mathrm{~atm}\right) .\) Calculate \(K\) at this temperature. What must the initial pressure of \(\mathrm{F}_{2}\) be to convert \(75.0 \%\) of the xenon to \(\mathrm{XeF}_{4}\) ?

Short answer

Also, what is the initial pressure of F2 required to achieve a 75% conversion of xenon to XeF4? The equilibrium constant (K) for the given reaction at the given temperature is 25. To achieve a 75% conversion of xenon to XeF4, the initial pressure of F2 must be approximately 0.346 atm.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is given by: $$ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g) $$

Calculate the equilibrium pressures of each species

50% of Xe reacts to form XeF4, which means the equilibrium pressure of Xe is 50% of the initial pressure: $$ P_{\mathrm{Xe}} = 0.20 \mathrm{~atm} \times 0.50 = 0.10 \mathrm{~atm} $$ Since the initial pressure of F2 is 0.40 atm and the stoichiometric ratio is 2 F2 : 1 Xe, the pressure change in F2 is twice the pressure change in Xe: $$ \Delta P_{\mathrm{F}_{2}} = 2 \times (0.20 - 0.10) \mathrm{~ at m} = 0.20 \mathrm{~atm} $$ Now, the equilibrium pressure of F2 can be calculated: $$ P_{\mathrm{F}_{2}} = 0.40 \mathrm{~atm} - 0.20 \mathrm{~atm} = 0.20 \mathrm{~atm} $$ Finally, the equilibrium pressure of XeF4, which is formed from 50% of the initial Xe, is: $$ P_{\mathrm{XeF}_{4}} =1-0.10 \mathrm{~atm} = 0.10 \mathrm{~atm} $$

Calculate the equilibrium constant K

Using the equilibrium pressures of each species, we can now calculate K using the following expression: $$ K = \frac{P_{\mathrm{XeF}_{4}}}{P_{\mathrm{Xe}} \times P_{\mathrm{F}_{2}}^2} $$ Plugging in the values, we get: $$ K = \frac{0.10}{0.10 \times 0.20^2} = 25 $$ Now, let's determine the initial pressure of F2 required to achieve a 75% conversion of xenon to XeF4.

Calculate the equilibrium pressures for 75% conversion of Xe

For 75% conversion of Xe, the equilibrium pressure of Xe is 75% of the initial pressure: $$ P_{\mathrm{Xe}} = 0.20 \mathrm{~atm} \times 0.25 = 0.05 \mathrm{~atm} $$ Since 75% of Xe is converted to XeF4, the equilibrium pressure of XeF4 is now: $$ P_{\mathrm{XeF}_{4}} = 0.20 \mathrm{~atm} - 0.05 \mathrm{~atm} = 0.15 \mathrm{~atm} $$

Calculate the initial pressure of F2 required for 75% conversion of Xe

Now, we rearrange the equation for K in terms of the initial pressure of F2 and plug in the known values: $$ P_{\mathrm{F}_{2}}^2 = \frac{P_{\mathrm{XeF}_{4}}}{K \times P_{\mathrm{Xe}}} $$ $$ P_{\mathrm{F}_{2}}^2 = \frac{0.15}{25 \times 0.05} = 0.12 $$ Finally, we take the square root of this result to find the initial pressure of F2: $$ P_{\mathrm{F}_{2}} = \sqrt{0.12} \mathrm{~atm} \approx 0.346 \mathrm{~atm} $$ So, in order to convert 75% of xenon to XeF4, the initial pressure of F2 must be approximately 0.346 atm.