Use the References to access important values if needed for this question. The equilibrium constant, Kp, for the following reaction is 1.80 x 10-2 at 698 K. 2HI(g) H₂(g) + 1₂ (9) If an equilibrium mixture of the three gases in a 11.2 L container at 698 K contains HI at a pressure of 1.13 atm and H₂ at a pressure of 1.03 atm, the equilibrium partial pressure of I₂ is atm. Submit Answer Retry Entire Group 9 more group attempts remaining

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The equilibrium constant, \( K_p \), for the following reaction is \( 1.80 \times 10^{-2} \) at 698 K.
\[ 2 \text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \]

If an equilibrium mixture of the three gases in a 11.2 L container at 698 K contains HI at a pressure of 1.13 atm and \(\text{H}_2\) at a pressure of 1.03 atm, the equilibrium partial pressure of \(\text{I}_2\) is \[ \underline{\phantom{aaa}} \] atm.

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Transcribed Image Text:The equilibrium constant, \( K_p \), for the following reaction is \( 1.80 \times 10^{-2} \) at 698 K. \[ 2 \text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \] If an equilibrium mixture of the three gases in a 11.2 L container at 698 K contains HI at a pressure of 1.13 atm and \(\text{H}_2\) at a pressure of 1.03 atm, the equilibrium partial pressure of \(\text{I}_2\) is \[ \underline{\phantom{aaa}} \] atm. **Options:** - Submit Answer - Retry Entire Group (9 more group attempts remaining) _Hint and navigation buttons:_ - Show Hint - Previous - Next **Note:** Background is provided by Cengage Learning | Cengage Technical Support.
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