Question
Go along with the James Madison High school Chemistry part 1 curriculum and find the equilibrium expression for the reaction: CaO(s)+CH4(g)+2H2O(g) ⇄ CaCO3(s)+4H2(g) kp=2344
Answers
GPT-4o mini
To find the equilibrium expression for the given reaction, we can follow the principles of chemical equilibrium. The equilibrium expression relates the concentrations (or partial pressures) of the reactants and products at equilibrium.
The given reaction is:
\[ \text{CaO(s)} + \text{CH}_4(g) + 2\text{H}_2\text{O(g)} \rightleftharpoons \text{CaCO}_3(s) + 4\text{H}_2(g) \]
**Step 1: Identify the phases of the reactants and products.**
- CaO and CaCO3 are solids (s).
- CH4 and H2O are gases (g).
- H2 is also a gas (g).
**Step 2: Write the equilibrium expression.**
The equilibrium expression is formulated using only the gases (reactants and products), because the concentrations of solids do not appear in the equilibrium expression (their activities are taken to be 1).
The general form for the equilibrium expression (Kp for gas-phase reactions) is given by:
\[
K_p = \frac{(P_{\text{products}})}{(P_{\text{reactants}})}
\]
**Step 3: Write the expression for the given reaction.**
For the reaction:
\[
K_p = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)}
\]
Since both CaO and CaCO3 are solids, they do not appear in the expression.
**Step 4: Substitute the known value for Kp.**
The given \( K_p = 2344 \), so we have:
\[
K_p = 2344 = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)}
\]
So, the equilibrium expression for the reaction \( \text{CaO(s)} + \text{CH}_4(g) + 2\text{H}_2\text{O(g)} \rightleftharpoons \text{CaCO}_3(s) + 4\text{H}_2(g) \) is:
\[
K_p = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)} = 2344
\]
This expression shows the relationship between the partial pressures of the gaseous reactants and products at equilibrium.
The given reaction is:
\[ \text{CaO(s)} + \text{CH}_4(g) + 2\text{H}_2\text{O(g)} \rightleftharpoons \text{CaCO}_3(s) + 4\text{H}_2(g) \]
**Step 1: Identify the phases of the reactants and products.**
- CaO and CaCO3 are solids (s).
- CH4 and H2O are gases (g).
- H2 is also a gas (g).
**Step 2: Write the equilibrium expression.**
The equilibrium expression is formulated using only the gases (reactants and products), because the concentrations of solids do not appear in the equilibrium expression (their activities are taken to be 1).
The general form for the equilibrium expression (Kp for gas-phase reactions) is given by:
\[
K_p = \frac{(P_{\text{products}})}{(P_{\text{reactants}})}
\]
**Step 3: Write the expression for the given reaction.**
For the reaction:
\[
K_p = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)}
\]
Since both CaO and CaCO3 are solids, they do not appear in the expression.
**Step 4: Substitute the known value for Kp.**
The given \( K_p = 2344 \), so we have:
\[
K_p = 2344 = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)}
\]
So, the equilibrium expression for the reaction \( \text{CaO(s)} + \text{CH}_4(g) + 2\text{H}_2\text{O(g)} \rightleftharpoons \text{CaCO}_3(s) + 4\text{H}_2(g) \) is:
\[
K_p = \frac{(P_{\text{H}_2})^4}{(P_{\text{CH}_4})(P_{\text{H}_2\text{O}}^2)} = 2344
\]
This expression shows the relationship between the partial pressures of the gaseous reactants and products at equilibrium.