A mixture of 1.60 moles of X2 gas and 2.70 moles of Y2 gas are mixed with 6.42 moles of XY gas in a 10.0 L tank at 25°C. X212) + Y 21g) ++2XY) Answer both questions: a. If the K, for this reaction is 2.4 x 10² at this temperature, is the reaction at equilibrium? b. If not, what is the direction of the net reaction?
A mixture of 1.60 moles of X2 gas and 2.70 moles of Y2 gas are mixed with 6.42 moles of XY gas in a 10.0 L tank at 25°C. X212) + Y 21g) ++2XY) Answer both questions: a. If the K, for this reaction is 2.4 x 10² at this temperature, is the reaction at equilibrium? b. If not, what is the direction of the net reaction?
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![**Problem Statement: Chemical Equilibrium Analysis**
A mixture of 1.60 moles of \( X_2 \) gas and 2.70 moles of \( Y_2 \) gas are mixed with 6.42 moles of \( XY \) gas in a 10.0 L tank at 25°C. The chemical reaction is as follows:
\[ X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)} \]
**Questions to Solve:**
a. If the \( K_c \) for this reaction is \( 2.4 \times 10^{-2} \) at this temperature, is the reaction at equilibrium?
b. If not, what is the direction of the net reaction?
**Analysis:**
1. **Determine Reaction Quotient (\( Q_c \))**:
- Calculate the concentrations of each species:
- \([X_2] = \frac{1.60 \, \text{moles}}{10.0 \, \text{L}}\)
- \([Y_2] = \frac{2.70 \, \text{moles}}{10.0 \, \text{L}}\)
- \([XY] = \frac{6.42 \, \text{moles}}{10.0 \, \text{L}}\)
2. **Reaction Quotient Expression**:
- \[ Q_c = \frac{[XY]^2}{[X_2][Y_2]} \]
3. **Comparison**:
- Compare \( Q_c \) to \( K_c \) to determine if the reaction is at equilibrium.
- If \( Q_c = K_c \), the reaction is at equilibrium.
- If \( Q_c < K_c \), the forward reaction is favored.
- If \( Q_c > K_c \), the reverse reaction is favored.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F82ecc600-dbac-4abc-88e0-4a90eb8088d9%2F69b93fa6-99f7-4ca3-9fc9-9d79bddba2ac%2Fowsx1jh_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Chemical Equilibrium Analysis**
A mixture of 1.60 moles of \( X_2 \) gas and 2.70 moles of \( Y_2 \) gas are mixed with 6.42 moles of \( XY \) gas in a 10.0 L tank at 25°C. The chemical reaction is as follows:
\[ X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)} \]
**Questions to Solve:**
a. If the \( K_c \) for this reaction is \( 2.4 \times 10^{-2} \) at this temperature, is the reaction at equilibrium?
b. If not, what is the direction of the net reaction?
**Analysis:**
1. **Determine Reaction Quotient (\( Q_c \))**:
- Calculate the concentrations of each species:
- \([X_2] = \frac{1.60 \, \text{moles}}{10.0 \, \text{L}}\)
- \([Y_2] = \frac{2.70 \, \text{moles}}{10.0 \, \text{L}}\)
- \([XY] = \frac{6.42 \, \text{moles}}{10.0 \, \text{L}}\)
2. **Reaction Quotient Expression**:
- \[ Q_c = \frac{[XY]^2}{[X_2][Y_2]} \]
3. **Comparison**:
- Compare \( Q_c \) to \( K_c \) to determine if the reaction is at equilibrium.
- If \( Q_c = K_c \), the reaction is at equilibrium.
- If \( Q_c < K_c \), the forward reaction is favored.
- If \( Q_c > K_c \), the reverse reaction is favored.
Expert Solution
Step 1
Given equilibrium is
X2(g) + Y2(g) <----> 2XY(g)
The expression for reaction quotient (Qc) for this reaction is
Qc = [XY]2 / [X2][Y2]
The direction of reaction can be predicted by comparing Qc value with equilibrium constant , Kc
- If Qc = Kc , reaction is at equilibrium
- If Qc < Kc , reaction will proceed in forward direction to achieve equilibrium
- If Qc > Kc , reaction will proceed in reverse direction to achieve equilibrium
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