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.
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.
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The equilibrium constant, Kp, for the following reaction is 1.80 x 10-2 at 698 K.
2HI(g) H₂(g) + 1₂ (9)
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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.
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The equilibrium constant, Kp, for the following reaction is 1.80 x 10-2 at 698 K.
2HI(g) H₂(g) + 1₂ (9)
80
Submit Answer
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.
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The equilibrium constant, Kp, for the following reaction is 0.160 at 298 K.
2NOBr(g) 2NO(g) + Br₂(g)
If an equilibrium mixture of the three gases in a 17.8 L container at 298 K contains NOBr at a pressure of 0.301 atm
and NO at a pressure of 0.376 atm, the equilibrium partial pressure of Br₂ is
atm.
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The equilibrium constant, Kp, for the following reaction is 0.160 at 298 K.
2NOBr(g) 2NO(g) + Br₂(g)
If an equilibrium mixture of the three gases in a 17.8 L container at 298 K contains NOBr at a pressure of 0.301 atm
and NO at a pressure of 0.376 atm, the equilibrium partial pressure of Br₂ is
atm.
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Expert Solution
Step 1
1) we have to calculate the equilibrium partial pressure of I2
2) we have to calculate the equilibrium partial pressure of Br2
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