Question

In: Chemistry

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 reactants and product when 0.388 moles of H2 and 0.388 moles of I2 are introduced into a 1.00 L vessel at 698 K. [ H2 ] = M [ I2 ] = M [ HI ] = M

Solutions

Expert Solution

1) The given reaction is,

_______PCl5(g) <--> PCl3(g)______ +_______ Cl2(g)

I ______0.279 _______0 ______________________0

C_________-x _______+x _____________________+x

E_______0.279 -x ______x _____________________x

Now,

Kc = [PCl3][Cl2]/[PCl5] = x * x / (0.279 - x) = 1.20 x 10-2

=> x2 / 0.279 - x = 1.20 x 10-2

=> x2 = (0.279 -x) 1.20 x 10-2

=> x2 = 0.279 x 1.20 x 10-2 - 1.20 x 10-2 * x

=> x2 +1.20 * 10-2 * x - 0.279 *1.20 * 10-2 = 0

=> x2 + 0.0120 x - 0.003348 = 0

Using quadratic formula , taking only posiitve value of x we get,

x = -0.0120 + 0.1163 / 2

=> x = 0.05215

[PCl3 ]= 0.05215 mol / 1 L = 0.05215 M

[Cl2] =  0.05215 mol / 1 L = 0.05215 M

and [PCl5] = 0.279 -0.05215 mol / 1 L = 0.22685 M

2) The given reaction is H2 (g) + I2 (g) <---- ----> 2 HI (g)

ICE table H2 (g) ______+_______ I2 (g) <---- ----> 2 HI (g)

I______0.388 ________________0.388 __________0

C_____-x _____________________-x____________+2x

E_____0.388 -x _______________0.388-x ________ 2x

Now, Kc = [HI]2 / [H2] [I2] = 55.6

=> 55.6 = (2x)2 / (0.388 -x )* (0.388 -x )

=> 55.6 = 4x2 / (0.388 -x)2

=> 55.6 * (0.388 -x)2 =4 x2

=> 55.6 *( 0.150544 - 0.776x + x2) = x2

=> 8.3702464 - 43.1456 x + 55.6x2 -4x2 = 0

=> 51.6 x2 -43.1456 x + 8.3702464 = 0

Using quadratic formula, we get two values of x as--

x = 43.1456 ( +-) 11.5725 / 103.2

=> x = 43.1456 + 11.5725/ 103.2 = 0.530 , or x = 43.1456 -11.5725 / 103.2 = 0.306

With 0.0.530 we egt negative value of concentration, so it cannot be used, therefore the accepted value of x will be 0.306

Now, [H2] = 0.388 - 0.306 mol / 1 L = 0.082 M

[I2] = 0.388 - 0.306 mol / 1 L = 0.082 M

[HI] = 2 x 0.306 mol / 1 L = 0.612 M

queen_honey_blossom answered 2 years ago
Next > < Previous

Related Solutions

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...
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)...
The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...
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 0.218 M PCl5, 5.11×10-2 M PCl3 and 5.11×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.92×10-2 mol of PCl3(g) is added to the flask? [PCl5] = M [PCl3] = M [Cl2] = M
The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...
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 0.200 M PCl5, 4.90×10-2 M PCl3 and 4.90×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.12×10-2 mol of Cl2(g) is added to the flask? [PCl5] = [PCl3] = [Cl2] =  
The equilibrium constant, Kc, for the following reaction is 1.05×10-3 at 446 K. PCl5(g) PCl3(g) +...
The equilibrium constant, Kc, for the following reaction is 1.05×10-3 at 446 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 446 K, the equilibrium concentration of Cl2(g) is found to be 0.498 M. Calculate the concentration of PCl5 in the equilibrium mixture. ____M _______________________________________________________________________________________________________________ A student ran the following reaction in the laboratory at 689 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 4.06×10-2 moles of N2(g) and 5.20×10-2...
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...
1) The equilibrium constant, K, for the following reaction is 2.02×10-2 at 513 K. PCl5(g) <------...
1) The equilibrium constant, K, for the following reaction is 2.02×10-2 at 513 K. PCl5(g) <------ ------> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 18.3 L container at 513 K contains 0.216 M PCl5, 6.60×10-2 M PCl3 and 6.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.25 L? [PCl5] = ______M [PCl3] =...
The equilibrium constant, K, for the following reaction is 1.87×10-2 at 511 K. PCl5(g) <--------->>>PCl3(g) +...
The equilibrium constant, K, for the following reaction is 1.87×10-2 at 511 K. PCl5(g) <--------->>>PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 13.1 L container at 511 K contains 0.209 M PCl5,   6.24×10-2 M PCl3 and 6.24×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 7.01 L? [PCl5] = M [PCl3] = M [Cl2] =...
The equilibrium constant, K, for the following reaction is 1.42×10-2 at 504 K. PCl5(g) --> PCl3(g)...
The equilibrium constant, K, for the following reaction is 1.42×10-2 at 504 K. PCl5(g) --> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 10.9 L container at 504 K contains 0.279 M PCl5, 6.29×10-2 M PCl3 and 6.29×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 4.95 L? [PCl5] = M [PCl3] = M [Cl2]...
The equilibrium constant, K, for the following reaction is 3.16×10-2 at 525 K. PCl5(g) <----->PCl3(g) +...
The equilibrium constant, K, for the following reaction is 3.16×10-2 at 525 K. PCl5(g) <----->PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 4.56 L container at 525 K contains 0.243 M PCl5,   8.77×10-2 M PCl3 and 8.77×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 10.3 L? [PCl5] = M [PCl3] = M [Cl2] = M
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT