Consider the following equilibrium for which H = -209.7: 2 C2H6(g) + O2(g) 2 C2H4(g) + 2 H2O(g) How will each of the following changes affect an equilibrium mixture of the 4 gases in this reaction?
Consider the following equilibrium for which H = -209.7:
2 C2H6(g) + O2(g) 2 C2H4(g) + 2 H2O(g)
How will each of the following changes affect an equilibrium mixture of the 4 gases in this reaction?
(a) C2H4(g) is added to the system.
-the equilibrium will not shift
-the equlibrium will shift toward product but Keq will not change
-the equlibrium will shift toward reactant but Keq will not change
-the equlibrium will shift toward product but Keq will increase
-the equlibrium will shift toward reactant but Keq will decrease
(b) The reaction mixture is heated.
-the equilibrium will not shift
-the equlibrium will shift toward product but Keq will not change
-the equlibrium will shift toward reactant but Keq will not change
-the equlibrium will shift toward product but Keq will increase
-the equlibrium will shift toward reactant but Keq will decrease
(c) The volume of the reaction vessel is reduced by 50%.
-the equilibrium will not shift
-the equlibrium will shift toward product but Keq will not change
-the equlibrium will shift toward reactant but Keq will not change
-the equlibrium will shift toward product but Keq will increase
-the equlibrium will shift toward reactant but Keq will decrease
(d) A catalyst is added to the reaction mixture.
-the equilibrium will not shift
-the equlibrium will shift toward product but Keq will not change
-the equlibrium will shift toward reactant but Keq will not change
-the equlibrium will shift toward product but Keq will increase
-the equlibrium will shift toward reactant but Keq will decrease
(e) The total pressure of the system is increased by adding a noble gas.
-the equilibrium will not shift
-the equlibrium will shift toward product but Keq will not change
-the equlibrium will shift toward reactant but Keq will not change
-the equlibrium will shift toward product but Keq will increase
-the equlibrium will shift toward reactant but Keq will decrease
(f) H2O(g) is removed from the system.
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![### Understanding Chemical Equilibrium Changes
**Equilibrium Reaction and Enthalpy:**
Consider the following equilibrium for which ΔH = -209.7 kJ:
\[ 2 \, \text{C}_2\text{H}_6 (\text{g}) + \text{O}_2 (\text{g}) \rightleftharpoons 2 \, \text{C}_2\text{H}_4\text{(g)} + 2 \, \text{H}_2\text{O(g)} \]
**How different changes affect an equilibrium mixture of the four gases in this reaction:**
**(a) Adding C\(_2\)H\(_4\)(g) to the system.**
- Incorrect Initial Response: The equilibrium will shift toward reactant and \(K_{eq}\) will decrease.
- Correct Response: The correct answer should consider Le Chatelier's principle, where adding a product shifts the equilibrium towards the reactants, but \(K_{eq}\) remains unchanged.
**(b) Heating the reaction mixture.**
- Correct Response: The equilibrium will shift toward reactants and \(K_{eq}\) will decrease.
- Explanation: Because the reaction is exothermic (\(\Delta H\) is negative), adding heat shifts the equilibrium to the left.
**(c) Reducing the volume of the reaction vessel by 50%.**
- Incorrect Initial Response: The equilibrium will shift toward reactants and \(K_{eq}\) will decrease.
- Correct Response: The reaction's response depends on the moles of gas. If pressure increases more on one side due to volume reduction, equilibrium shifts to the side with fewer gas moles.
**(d) Adding a catalyst to the reaction mixture.**
- Correct Response: The equilibrium will not shift.
- Explanation: A catalyst speeds up both the forward and reverse reactions equally, without affecting the position of equilibrium.
**(e) Increasing the total pressure of the system by adding a noble gas.**
- Correct Response: The equilibrium will not shift.
- Explanation: Adding inert gas increases overall pressure but does not affect partial pressures of reactants/products, so equilibrium remains unchanged.
### Summary
This exercise demonstrates Le Chatelier’s principle and how different factors affect chemical equilibrium, particularly the position of equilibrium and equilibrium constant \(K_{eq}\).](https://content.bartleby.com/qna-images/question/ee401d09-5414-49af-9e22-dedff044897d/099cc662-e1f4-4e1e-a2a4-41238ee101cc/vlawji_processed.jpeg)
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