Calculate a value for the equilibrium constant for the reaction O2 (9) + O(g) = 03 (g) given hv NO2 (g) = NO(g) + O(g) K = 5.1 × 10-49 03 (g) + NO(9) = NO,(g)+ O2(9) K= 2.2 × 10-34 (Hint: When reactions are added together, the equilibrium expressions are multiplied.) K=

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**Calculating the Equilibrium Constant**

Calculate a value for the equilibrium constant for the reaction:

\[ \text{O}_2(g) + \text{O}(g) \rightleftharpoons \text{O}_3(g) \]

Given:

1. \[ \text{NO}_2(g) \xrightleftharpoons{h\nu} \text{NO}(g) + \text{O}(g) \quad K = 5.1 \times 10^{-49} \]

2. \[ \text{O}_3(g) + \text{NO}(g) \rightleftharpoons \text{NO}_2(g) + \text{O}_2(g) \quad K = 2.2 \times 10^{-34} \]

*(Hint: When reactions are added together, the equilibrium expressions are multiplied.)*

\[ K = \text{(resulting equilibrium constant)} \]
Transcribed Image Text:**Calculating the Equilibrium Constant** Calculate a value for the equilibrium constant for the reaction: \[ \text{O}_2(g) + \text{O}(g) \rightleftharpoons \text{O}_3(g) \] Given: 1. \[ \text{NO}_2(g) \xrightleftharpoons{h\nu} \text{NO}(g) + \text{O}(g) \quad K = 5.1 \times 10^{-49} \] 2. \[ \text{O}_3(g) + \text{NO}(g) \rightleftharpoons \text{NO}_2(g) + \text{O}_2(g) \quad K = 2.2 \times 10^{-34} \] *(Hint: When reactions are added together, the equilibrium expressions are multiplied.)* \[ K = \text{(resulting equilibrium constant)} \]
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