1. A flask contains 2.000 atm H2, 0.2056 atm I2, and 3.009 atm HI. The equilibrium constant is 92.6. If the reaction must proceed towards the right, calculate the equilibrium pressures.
H2 (g)+I2 (g)⇌2HI(g)
Answers
The equilibrium reaction is shown below.
The equilibrium reaction is shown below.2HI(g)⇌H
The equilibrium reaction is shown below.2HI(g)⇌H 2
The equilibrium reaction is shown below.2HI(g)⇌H 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HI
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.2
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.20
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2x
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xx
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxx
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2x
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xx
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxx
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04Hence, 2x=0.2−0.04=0.16 or x=0.08.
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04Hence, 2x=0.2−0.04=0.16 or x=0.08.The expression for the equilibrium constant is K
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04Hence, 2x=0.2−0.04=0.16 or x=0.08.The expression for the equilibrium constant is K P
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04Hence, 2x=0.2−0.04=0.16 or x=0.08.The expression for the equilibrium constant is K P
The equilibrium reaction is shown below.2HI(g)⇌H 2 (g)+I 2 (g)HIH 2 I 2 Initial0.200Change-2xxxEquilibrium0.2-2xxxBut the equilibrium pressure of HI is 0.04 atm.0.2−2x=0.04Hence, 2x=0.2−0.04=0.16 or x=0.08.The expression for the equilibrium constant is K P