Question

1) The equilibrium constant, K_{p}, for the following
reaction is **0.497** at **500** K:

**PCl _{5}**(g) <---
----->

Calculate the equilibrium partial pressures of all species when

P_{PCl5} |
= | _____atm |

P_{PCl3} |
= | ______atm |

P_{Cl2} |
= | ______atm |

2) The equilibrium constant, K_{p}, for the following
reaction is **55.6** at **698** K:

**H _{2}**(g) +

Calculate the equilibrium partial pressures of all species when

P_{H2} |
= | _____atm |

P_{I2} |
= | _____atm |

P_{HI} |
= | ______atm |

Answer #1

Kp = 0.497

Kp = PCl3*CL2/PCl5

initially

PCl5 = 1.52

PCl3 = 0

Cl2 = 0

in equilbirium

PCl5 = 1.52 - x

PCl3 = 0 + x

Cl2 = 0 + x

substitute

Kp = PCl3*CL2/PCl5

0.497= x*x/(1.52 - x)

x^2 + (0.497)x - (1.52*0.497) = 0

x = 0.6554

PCl5 = 1.52 - 0.6554 = 0.8646 atm

PCl3 = 0 + 0.6554 atm

Cl2 = 0 + 0.6554 atm

Q2

Kp = 55.6

Pinitial:

H2 = 2.18

I2 = 2.18

HI = 0

in equilbirium

H2 = 2.18 - x

I2 = 2.18 - x

HI = 0 +2x

Kp = HI^2 / (H2)(I2)

55.6 = (2x)^2 / (2.18 - x)(2.18 - x)

sqrt(55.6) = 2x/(2.18 - x)

2/7.456x = 2.18-x

(2/7.456 +1)x = 2.18

x = 2.18/((2/7.456 +1)) = 1.7189

H2 = 2.18 - 1.7189 = 0.4611 atm

I2 = 2.18 - 1.7189 = 0.4611 atm

HI = 0 +2*1.7189 = 3.4378 atm

The equilibrium constant, Kp, for the following reaction is
9.52×10-2 at 350 K:
CH4(g) + CCl4(g) 2CH2Cl2(g)
Calculate the equilibrium partial pressures of all species when
CH4 and CCl4, each at an intitial partial pressure of 1.12 atm, are
introduced into an evacuated vessel at 350 K.
PCH4 = ________ atm
PCCl4 = ________atm
PCH2Cl2 = ________ atm

The equilibrium constant, Kp, for the following reaction is 10.5
at 350 K:
2CH2Cl2(g) CH4(g) + CCl4(g)
Calculate the equilibrium partial pressures of all species when
CH2Cl2(g) is introduced into an evacuated flask at a pressure of
0.856 atm at 350 K.
PCH2Cl2 = ________ atm
PCH4 = _________ atm
PCCl4 = _______ atm

The equilibrium constant Kp for the reaction
C(s)+H2O(g)?CO(g)+H2(g) is 2.44
at 1000 K. What are the equilibrium partial pressures of H2O, CO,
and H2 if the initial partial pressures are PCO= 1.25 atm,
and PH2= 1.60 atm?
What is the equilibrium partial pressure of H2O?
What is the equilibrium partial pressure of CO?
What is the equilibrium partial pressure of H2?

Consider the following reaction: H2(g)+I2(g)?2HI(g) A reaction
mixture at equilibrium at 175 K contains PH2=0.958atm,
PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at
175 K, contains PH2=PI2= 0.630 atm , and PHI= 0.102 atm .
Is the second reaction at equilibrium?
If not, what will be the partial pressure of HI when the
reaction reaches equilibrium at 175 K?

Consider the reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K contains
PH2=0.958atm, PI2=0.877atm,
and PHI=0.020atm. A second reaction mixture,
also at 175 K, contains
PH2=PI2=0.629
atm , and PHI= 0.101 atm .
Is the second reaction at equillibrium? I found that the
kp=4.76X10-4 and Qp = .0257 So no, not at
equillibrium.
If not, what is the partial pressure of HI when the
reaction reaches equilibrium at 175 K? I need help
figuring out the ICE chart and...

Consider the following reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K
containsPH2=0.958atm, PI2=0.877atm, and
PHI=0.020atm. A second reaction mixture, also at 175 K,
contains PH2=PI2= 0.628 atm , and PHI=
0.107 atm .
A) If not, what will be the partial pressure of HI when the
reaction reaches equilibrium at 175 K?

For the reaction below, Kp= 28.09 at 800K. Calculate the
equilibrium partial pressure of the reactants and products if the
intitial pressure arePpCl5= 0.5300 atm and PpCl3= .5200 atm.
PCl5--> PCl3 + Cl2
PCl5=
Cl2=
PCl3=

The equilibrium constant, Kc, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) Calculate the
equilibrium concentrations of reactant and products when 0.390
moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K.

For the reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 70.5°C, Kp =
1.05. If one starts with 1.80 atm pressure of PCl3(g), 1.72 atm
pressure of Cl2(g), and no PCl5(g), what is the partial pressure of
PCl5(g) at equilibrium?
The answer is 0.856 atm
can you explain me how to solve this question?

For the exothermic reaction
PCl3(g)+Cl2(g)?PCl5(g)
Kp = 0.200 at a certain temperature.
A flask is charged with 0.500 atm PCl3 , 0.500
atm Cl2, and 0.300 atm PCl5 at this temperature.
What are the equilibrium partial pressures of PCl3
, Cl2, and PCl5, respectively?
Express your answers numerically in atmospheres with three
digits after the decimal point, separated by commas.

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