Question

In: Chemistry

1. An 0.865-mol sample of PCl5 is placed in a 500.-mL reaction vessel. What is the...

1. An 0.865-mol sample of PCl5 is placed in a 500.-mL reaction vessel. What is the concentration of each substance when the reaction PCl5(g) PCl3(g) + Cl2(g) has reached equilibrium at 250 deg Celcius (when Kc = 1.80)?

Solutions

Expert Solution

                PCl5(g) ------------>PCl3(g) + Cl2(g)

I              0.865                       0               0

C              -x                            +x                +x

E            0.865-x                        +x              +x

                [PCl5]   = 0.865-x/0.5

               [PCl3]    = x/0.5

              [Cl2]        = x/0.5

          Kc   = [PCl3][Cl2]/[PCl5]

         1.8    = x*x/0.5*(0.865-x)

         1.8*0.5*(0.865-x) = x^2

              x    = 0.54

[PCl5]   = 0.865-x/0.5    = 0.865-0.54/0.5   = 0.325/0.5   = 0.65M

               [PCl3]    = x/0.5    = 0.54/0.5            = 1.08M

              [Cl2]        = x/0.5    = 0.54/0.5            = 1.08M

         


Related Solutions

2.0 mol each of PCl5, PCl3, and Cl2 are placed in a 2.0L reaction vessel at...
2.0 mol each of PCl5, PCl3, and Cl2 are placed in a 2.0L reaction vessel at 620K. Kc for the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) is 0.60 at the given temperature. What is the equilibrium concentration of Cl2? 1.146 is incorrect.
A 0.268 mol sample of PCl5(g) is injected into an empty 4.05 L reaction vessel held...
A 0.268 mol sample of PCl5(g) is injected into an empty 4.05 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium. 1.80 @ 250 C
Consider the following reaction: PCl5(g)⇌PCl3(g)+Cl2(g) Initially, 0.65 mol of PCl5 is placed in a 1.0 L...
Consider the following reaction: PCl5(g)⇌PCl3(g)+Cl2(g) Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.16 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5? Express your answer to two significant figures and include the appropriate units.
A 2.00 mol sample of nitrogen dioxide was placed in a 40.0 L vessel. Nitrogen dioxide...
A 2.00 mol sample of nitrogen dioxide was placed in a 40.0 L vessel. Nitrogen dioxide decomposes according to the equation At a certain temperature, the nitrogen dioxide was 8.6%; that is, for each mol of NO2 before the reaction, (1.000 – 8.6×10-2) mol NO2 remains after dissociation. Calculate the value of KC for this reaction.
A 7.95-atm PCl5 is initially placed in an empty vessel at 25˚C. Calculate Kp if the...
A 7.95-atm PCl5 is initially placed in an empty vessel at 25˚C. Calculate Kp if the equilibrium partial pressure of Cl2 is 3.00 atm at the same temperature. I am completely lost on how to even start the problem, let alone solve it.
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?
1)The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) The equilibrium...
1)The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <------ ------> PCl3(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.14 atm at 500 K. PPCl5 = atm PPCl3 = atm PCl2 = atm 2) The equilibrium constant, Kp, for the following reaction is 55.6 at 698...
When CS2(g) (6.828 mol) and 27.31 mol of H2(g) in a 360.0 L reaction vessel at...
When CS2(g) (6.828 mol) and 27.31 mol of H2(g) in a 360.0 L reaction vessel at 382.0 °C are allowed to come to equilibrium the mixture contains 52.58 grams of CH4(g). What concentration (mol/L) of H2(g) reacted?   CS2(g)+4H2(g) = CH4(g)+2H2S(g) When SbCl3(g) (4.334 mol) and 307.3 grams of Cl2(g) in a 420.0 L reaction vessel at 716.0 °C are allowed to come to equilibrium the mixture contains 0.007729 mol/L of SbCl5(g). What is the equilibrium concentration (mol/L) of SbCl3(g)? SbCl3(g)+Cl2(g)...
1) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <-----...
1) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <----- ------> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.201 M PCl5, 4.91×10-2 M PCl3 and 4.91×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.27×10-2 mol of Cl2(g) is added to the flask? [PCl5] = M [PCl3] = M [Cl2] = M 2)...
1) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <---...
1) 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 0.279 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K. [PCl5] = M [PCl3] = M [Cl2] = M 2) The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K. H2 (g) + I2 (g) <---- ----> 2 HI (g) Calculate the equilibrium concentrations of...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT