Question

In: Chemistry

For the exothermic reaction PCl3(g)+Cl2(g)?PCl5(g) Kp = 0.200 at a certain temperature. A flask is charged...

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.

Expert Solution

For the reaction,

Kp = 0.200

initial,

PCl3 = Cl2 = 0.5 atm

PCl5 = 0.3 atm

ICE chart

PCl3(g) + Cl2(g) <==> PCl5(g)

I 0.5 0.5 0.3

C -x -x +x

E 0.5-x 0.5-x 0.3+x

So,

Kp = [PCl5]/[PCl3][Cl2]

0.200 = (0.3+x)/(0.5-x)(0.5-x)

0.05 - 0.2x + 0.2x^2 = 0.3 + x

0.2x^2 - 1.2x - 0.25 = 0

x = -0.2025 atm

at equilibrium,

partial pressure of PCl3 = 0.5 + 0.2025 = 0.702 atm

partial pressure of Cl2 = 0.5 + 0.2025 = 0.702 atm

partial pressure of PCl5 = 0.3 - 0.2025 = 0.097 atm

queen_honey_blossom answered 1 year ago

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.140 at a certain temperature. A flask is charged...

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.140 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.

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.160 at a certain temperature. A flask is charged...

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.160 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? How will the following changes affect the mole fraction of chlorine gas, X Cl2, in the equilibrium mixture. X Cl2 increases when __________ X Cl2 decreases when _________ X Cl2 stays the same __________ Fill in the blanks...

Q2. For the exothermic reaction PCl3(g)+Cl2(g) <======>PCl5(gas) Kp=0.160 at a certian temperature. A flask is charged...

Q2. For the exothermic reaction PCl3(g)+Cl2(g) <======>PCl5(gas) Kp=0.160 at a certian temperature. A flask is charged with 0.500 atm PCl3, 0.500 atm Cl2, and .300 atm PCl5 at this temperature. True or False __________The net reaction proceeds to the left to attain equlibrium __________Q is less than k. __________Q is equal to k. __________The reaction is at equlibrium __________Q is greater than k. ___________ No net reaction will occur ____________ The net reaction proceeds to the right to attain equlibrium....

The reaction below is exothermic: PCl3 (g) + Cl2 (g) <---> PCl5 (g) Which change will...

The reaction below is exothermic: PCl3 (g) + Cl2 (g) <---> PCl5 (g) Which change will increase the number of moles of of PCl5 (g) present? A. The volume of the reaction vessel is tripled B. The reaction vessel is cooled C. Some of the Cl2 is removed D. Krypton gas is added to the reaction vessel

1.For the reaction, PCl5(g) <-----> PCl3(g) + Cl2(g) Kp = 24.6 at 500 K calculate the equilibrium...

1.For the reaction, PCl5(g) <-----> PCl3(g) + Cl2(g) Kp = 24.6 at 500 K calculate the equilibrium partial pressures of the reactants and products if the initial pressures are PPCl5 = 0.610 atm, PPCl3 = 0.400 atm and PCl2 = 0.000 atm. PPCl5 = PPCl3 = PCl2 = 2. H2O(g) + Cl2O(g) <-----> 2HClO(g) Kc = 0.14 at 298.15 K calculate the equilibrium concentrations of the reactants and products if the initial concentrations are [H2O(g)] = 0.00482 mol L-1, [Cl2O(g)] = 0.00482...

Given the equation: PCl3(g) + Cl2(g) ⇌ PCl5(g) kp = 1.05 at 70.5 °C If one...

Given the equation: PCl3(g) + Cl2(g) ⇌ PCl5(g) kp = 1.05 at 70.5 °C If one starts with 1,7 atm of PCl5(g), what is the partial pressure of PCl5(g) at equilibrium?

The reaction: PCl5(g) <-> PCl3(g) + Cl2(g) has Kc=0.0900. A 0.1000 mol sammple of PCl5 is...

The reaction: PCl5(g) <-> PCl3(g) + Cl2(g) has Kc=0.0900. A 0.1000 mol sammple of PCl5 is placed in an empty 1.00 flask and the above reaction is allowed to come to equilibrium at a certain temp. How manny moles of PCl5, PCl3 and Cl2, respectively, are present at equilibrium?

Consider the following reaction: PCl5 (g)⇄ PCl3 (g) + Cl2 (g) ΔH = 87.9 kJ/reaction For...

Consider the following reaction: PCl5 (g)⇄ PCl3 (g) + Cl2 (g) ΔH = 87.9 kJ/reaction For each of the following changes, indicate whether the reaction will proceed towards products, towards reactants, or have no net reaction to reestablish equilibrium. a. ___________ Temperature is increased b. ___________ The pressure is increased by decreasing the volume. c. ___________ The pressure is decreased by removing some PCl 5 (g) d. ___________ The pressure is increased by adding N2(g) e. ___________ Cl2(g) is removed...

What is the percent yield for the reaction PCl3(g) + Cl2(g) --> PCl5(g) if 119.3 g...

What is the percent yield for the reaction PCl3(g) + Cl2(g) --> PCl5(g) if 119.3 g of PCl5 are formed when 61.3 g of Cl2 reacts with 125.5g PCl3?

1) At a certain temperature, 322 K, Kp for the reaction, Cl2(g) <=> 2 Cl(g), is...

1) At a certain temperature, 322 K, Kp for the reaction, Cl2(g) <=> 2 Cl(g), is 7.82 x 10-60. Calculate the value of ΔGo in kJ for the reaction at this temperature. 2) The equilibrium constant for the reaction, 2 Fe3+(aq) + Hg22+(aq) <=> 2 Fe2+(aq) + 2 Hg2+(aq) is 9.1 x 10-6(aq) at 298 K. Calculate ΔG in J when {Fe3+(aq)} = 0.375 {Hg22+(aq)} = 0.051 {Fe2+(aq)} = 0.021 {Hg2+(aq)} = 0.06