Asked by Brooke
Consider the following reaction:
H2+I2<->2HI
A reaction mixture at equilibrium at 175 K contains P(H2)=0.958 atm, P(I2)=0.877 atm, and P(HI)=0.020 atm. A second reaction mixture, also at 175 K, contains P(H2)=P(I2)=0.616 atm, and P(HI)=0.106 atm. What will the partial pressure of HI be when the reaction reaches equilibrium at 175 K?
H2+I2<->2HI
A reaction mixture at equilibrium at 175 K contains P(H2)=0.958 atm, P(I2)=0.877 atm, and P(HI)=0.020 atm. A second reaction mixture, also at 175 K, contains P(H2)=P(I2)=0.616 atm, and P(HI)=0.106 atm. What will the partial pressure of HI be when the reaction reaches equilibrium at 175 K?
Answers
Answered by
DrBob222
1.Use the first set of conditions to calculate K for the reaction.
2. I would calculate reaction Q next with the second set of conditions to see which way the reaction will proceed to reach equilibrium.
3. Set up an ICE chart, after step 2, and substitute the values into K.
2. I would calculate reaction Q next with the second set of conditions to see which way the reaction will proceed to reach equilibrium.
3. Set up an ICE chart, after step 2, and substitute the values into K.
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