A student ran the following reaction in the laboratory at 445 K : PCl5(g)-------> PCl3(g) +...

A student ran the following reaction in the laboratory at 445 K

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

When she introduced 1.28 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of PCl5(g) to be 1.24 M.

Calculate the equilibrium constant, Kc, she obtained for this reaction.

Solutions

Expert Solution

molarity of PCl5 = no of moles/volume in L

= 1.28/1 = 1.28M

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

I 1.28 0 0

C -0.04 0.04 0.04

E 1.24 0.04 0.04

[PCl5] = 1.24M

[PCl3] = 0.04M

[Cl2] = 0.04M

Kc = [PCl3][Cl2]/[PCl5]

= 0.04*0.04/1.24 = 0.0013 = 1.3*10^-3 >>>>>answer

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

1. A student ran the following reaction in the laboratory at 582 K: CO(g) + Cl2(g)...

1. A student ran the following reaction in the laboratory at 582 K: CO(g) + Cl2(g) COCl2(g) When she introduced 0.380 moles of CO(g) and 0.403 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of COCl2(g) to be 0.342 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. 2. The equilibrium constant, Kc, for the following reaction is 1.29×10-3 at 540 K. COCl2(g) CO(g) + Cl2(g) When a sufficiently large sample of COCl2(g)...

A student ran the following reaction in the laboratory at 684 K: N2(g) + 3H2(g) 2NH3(g)...

A student ran the following reaction in the laboratory at 684 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 3.26×10-2 moles of N2(g) and 6.07×10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 5.83×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = __

-A student ran the following reaction in the laboratory at 1143 K: 2SO2(g) + O2(g) 2SO3(g)...

-A student ran the following reaction in the laboratory at 1143 K: 2SO2(g) + O2(g) 2SO3(g) When she introduced 8.19×10-2 moles of SO2(g) and 8.56×10-2 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of O2(g) to be 6.08×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =____________ -A student ran the following reaction in the laboratory at 363 K: CH4(g) + CCl4(g) 2CH2Cl2(g) When she introduced 4.64×10-2 moles of CH4(g) and...

A student ran the following reaction in the laboratory at 720 K: H2(g) + I2(g) 2HI(g)...

A student ran the following reaction in the laboratory at 720 K: H2(g) + I2(g) 2HI(g) When she introduced 0.189 moles of H2(g) and 0.218 moles of I2(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 6.05×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

A student ran the following reaction in the laboratory at 291 K: 2CH2Cl2(g) CH4(g) + CCl4(g)...

A student ran the following reaction in the laboratory at 291 K: 2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.63×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CH2Cl2(g) to be 4.92×10-3 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = _____

A student ran the following reaction in the laboratory at 610 K: CO(g) + Cl2(g) COCl2(g)...

A student ran the following reaction in the laboratory at 610 K: CO(g) + Cl2(g) COCl2(g) When she introduced 0.183 moles of CO(g) and 0.211 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.72×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. 2.A student ran the following reaction in the laboratory at 546 K: COCl2(g) CO(g) + Cl2(g) When she introduced 0.854 moles of COCl2(g) into a 1.00...

A student ran the folllowing reaction in the laboratory at 661 K: 2NH3(g) N2(g) + 3H2(g)...

A student ran the folllowing reaction in the laboratory at 661 K: 2NH3(g) N2(g) + 3H2(g) When she introduced NH3(g) at a pressure of 0.597 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of H2(g) to be 0.879 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.

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

Subjects