5) An equilibrium mixture of CO, O2, and CO₂ at 298 K contains 1.0 x 10-² atm CO2, 8.4 x 10-3 atm CO, 1.0 x 10-1 atm O2. Calculate the equilibrium constant (Kp) for the reaction below at 298 K. 2 CO(g) + O2(g) Answer: 14 → 2 CO2 (g)
5) An equilibrium mixture of CO, O2, and CO₂ at 298 K contains 1.0 x 10-² atm CO2, 8.4 x 10-3 atm CO, 1.0 x 10-1 atm O2. Calculate the equilibrium constant (Kp) for the reaction below at 298 K. 2 CO(g) + O2(g) Answer: 14 → 2 CO2 (g)
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Chapter1: Chemical Foundations
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![**Calculating the Equilibrium Constant (Kp) for a Chemical Reaction**
**Problem Statement:**
An equilibrium mixture of CO, O₂, and CO₂ at 298 K contains:
- CO₂: 1.0 x 10⁻² atm
- CO: 8.4 x 10⁻³ atm
- O₂: 1.0 x 10⁻¹ atm
Calculate the equilibrium constant (Kp) for the reaction below at 298 K.
**Reaction:**
\[ 2 \text{CO(g)} + \text{O}_2 \text{(g)} \rightleftharpoons 2 \text{CO}_2 \text{(g)} \]
**Answer:**
Kp = 14
**Explanation:**
- The problem involves calculating the equilibrium constant (Kp) for a reaction involving carbon monoxide (CO), oxygen (O₂), and carbon dioxide (CO₂).
- The equilibrium constant is calculated using the partial pressures of the gases at equilibrium.
- Kp is a ratio of the products' partial pressures raised to the power of their stoichiometric coefficients over the reactants’ partial pressures raised to their respective stoichiometric coefficients.
In this case:
\[ K_p = \frac{(\text{P}_{\text{CO}_2})^2}{(\text{P}_{\text{CO}})^2 (\text{P}_{\text{O}_2})} = \frac{(1.0 \times 10^{-2})^2}{(8.4 \times 10^{-3})^2 \times (1.0 \times 10^{-1})} \]
The given answer is 14.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1077e877-4e25-45d6-8c49-9aaf109d8b6e%2Fda2fb5a6-10e2-465f-8b5e-cea231280f04%2Fe0jkmob_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating the Equilibrium Constant (Kp) for a Chemical Reaction**
**Problem Statement:**
An equilibrium mixture of CO, O₂, and CO₂ at 298 K contains:
- CO₂: 1.0 x 10⁻² atm
- CO: 8.4 x 10⁻³ atm
- O₂: 1.0 x 10⁻¹ atm
Calculate the equilibrium constant (Kp) for the reaction below at 298 K.
**Reaction:**
\[ 2 \text{CO(g)} + \text{O}_2 \text{(g)} \rightleftharpoons 2 \text{CO}_2 \text{(g)} \]
**Answer:**
Kp = 14
**Explanation:**
- The problem involves calculating the equilibrium constant (Kp) for a reaction involving carbon monoxide (CO), oxygen (O₂), and carbon dioxide (CO₂).
- The equilibrium constant is calculated using the partial pressures of the gases at equilibrium.
- Kp is a ratio of the products' partial pressures raised to the power of their stoichiometric coefficients over the reactants’ partial pressures raised to their respective stoichiometric coefficients.
In this case:
\[ K_p = \frac{(\text{P}_{\text{CO}_2})^2}{(\text{P}_{\text{CO}})^2 (\text{P}_{\text{O}_2})} = \frac{(1.0 \times 10^{-2})^2}{(8.4 \times 10^{-3})^2 \times (1.0 \times 10^{-1})} \]
The given answer is 14.
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