Question:

The equilibrium constant for the reaction \[ 3 \, \text{NO} (g) \rightleftharpoons \text{N}_2\text{O} (g) + \text{NO}_2 (g) \] at 25°C is closest to: \[ [\Delta G^\circ = -104.18 \, \text{kJ}; R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}] \]

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When calculating the equilibrium constant from Gibbs free energy, use the equation \( \Delta G^\circ = -RT \ln K \).
Updated On: Jan 7, 2026
  • 1.043
  • \(1.8 \times 10^{18}\)
  • 1.651
  • \(5.7 \times 10^{-19}\)
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The Correct Option is B

Solution and Explanation

We use the relationship between Gibbs free energy and equilibrium constant: \[ \Delta G^\circ = -RT \ln K \] Substitute the given values into the equation: \[ -104.18 \times 10^3 = - (8.314)(298) \ln K \] Solving for \( K \), we find: \[ K = \exp\left(\frac{-104.18 \times 10^3}{-8.314 \times 298}\right) \approx 1.8 \times 10^{18} \] Final Answer: \(1.8 \times 10^{18}\)
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