This question is about chemical equilibria.
Give two essential features of a reaction in dynamic equilibrium.
A homogeneous gas-phase reaction is at equilibrium in a sealed container. When the pressure is increased, the equilibrium yield of the product increases. State what can be deduced about the chemical equation for this equilibrium.
Xenon and fluorine react to form xenon tetrafluoride according to the following equation: Xe(g)+2F2(g)⇌XeF4(g)\text{Xe(g)} + 2\text{F}_2\text{(g)} \rightleftharpoons \text{XeF}_4\text{(g)}Xe(g)+2F2(g)⇌XeF4(g)
0.800 mol0.800\text{ mol}0.800 mol of xenon is mixed with 1.60 mol1.60\text{ mol}1.60 mol of fluorine. At equilibrium, the total pressure in the vessel is 400 kPa400\text{ kPa}400 kPa and the mixture contains 0.200 mol0.200\text{ mol}0.200 mol of xenon tetrafluoride.
Calculate the amount, in moles, of xenon present at equilibrium. Then, calculate the partial pressure, in kPa\text{kPa}kPa, of xenon in this equilibrium mixture.
Give an expression for the equilibrium constant (KpK_{\text{p}}Kp) for this reaction: Xe(g)+2F2(g)⇌XeF4(g)\text{Xe(g)} + 2\text{F}_2\text{(g)} \rightleftharpoons \text{XeF}_4\text{(g)}Xe(g)+2F2(g)⇌XeF4(g)
A different mixture of xenon and fluorine is allowed to reach equilibrium at a temperature TTT. Some data for this equilibrium are shown in Table 1 below:
Table 1
| Substance | Equilibrium value |
|---|---|
| Partial pressure of Xe\text{Xe}Xe | 125 kPa125\text{ kPa}125 kPa |
| Partial pressure of XeF4\text{XeF}_4XeF4 | 5.00 kPa5.00\text{ kPa}5.00 kPa |
| KpK_{\text{p}}Kp | 6.40×10−7 kPa−26.40 \times 10^{-7}\text{ kPa}^{-2}6.40×10−7 kPa−2 |
Calculate the partial pressure, in kPa\text{kPa}kPa, of fluorine in this equilibrium mixture.
Use the KpK_{\text{p}}Kp value from Table 1 to calculate the value of KpK_{\text{p}}Kp for the following reaction at temperature TTT: XeF4(g)⇌Xe(g)+2F2(g)\text{XeF}_4\text{(g)} \rightleftharpoons \text{Xe(g)} + 2\text{F}_2\text{(g)}XeF4(g)⇌Xe(g)+2F2(g)
Give the units for this KpK_{\text{p}}Kp.