PCl5 decomposes to PCl3 and Cl2 with kp = 3.70*10^-6 at 975 K. Given that delta H run = -74.0 Kj/mol determine temperature at which Kp = 5.20*10^-4
PCl5 decomposes to PCl3 and Cl2 with kp = 3.70*10^-6 at 975 K. Given that delta...
For the reaction PCl5(g)<->PCl3(g)+Cl2(g) delta H=+93.6kJ and K=11.5 at 600 Kelvin. If K=1.2*10 at -4 power. Based on that data, when K=1.2*10 at -4 power is the temperature greater than or less than 600 Kelvin? Briefly explain your answer.
At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?
Phosphorus pentachloride decomposes at higher temperatures. PCl5(g) ⇄ PCl3(g) + Cl2(g) An equilibrium mixture at some temperature consists of 5.93 g PCl5, 208.23 g/mol 4.86 g PCl3, 137.33 g/mol 3.59 g Cl2, 70.91 g/mol in a 1.00-L flask. If you add 1.31 g of Cl2, how will the equilibrium be affected and what will the concentration of PCl5 be when equilibrium is reestablished? _shift left _shift right _no shift will occur [PCl5] =? mol/L
Phosphorus pentachloride decomposes according to the chemical equation PCl5(g)↽−−⇀PCl3(g)+Cl2(g)?c=1.80 at 250° C A 0.268 mol sample of PCl5(g) is injected into an empty 3.70 L reaction vessel held at 250° C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
Suppose K= 9.4 x 10^-3 at a certain temperature for the reaction PCl5--->PCl3+Cl2. Iif it is found that the concentration of PCl5 is twice the concentration of PCl3, what must be the concentration of Cl2 under these conditions?
Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = + 157 kJ P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = - 1207 kJ What is the standard-state enthalpy change for the following reaction? P4(g) + 10 Cl2(g) → 4PCl5(s)a -2100 kJb-1835 kJc-1364 kJD -1786 kJ
For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) PCl3(g) + Cl2(g) Suppose that 2.210 g of PCl5 is placed in an evacuated 535 mL bulb, which is then heated to 600. K. (a) What would be the pressure of PCl5 if it did not dissociate? (b) What is the partial pressure of PCl5 at equilibrium? (c) What is the total pressure in the bulb at equilibrium? (d) What is the degree of dissociation of...
QUESTION 1 In the reaction: PCl5 <----> PCl3 + Cl2 When the amount of PCl5 is increased, which of the following will occur? Check all that apply, deductions for incorrect choices. 1. the reaction will move to the right 2. The reaction will move to the left 3. PCl3 will increase 4. PCl3 will decrease 5. Cl2 will increase 6. Cl2 will decrease QUESTION 2 In the reaction: PCl5 <----> PCl3 + Cl2 When the amount of PCl3 is increased,...
The equilibrium constant for the following reaction is 1.42×10-3 at 179 °C. PCl5(g)<-->PCl3(g) + Cl2(g) K = 1.42×10-3 at 179 °C Calculate the equilibrium constant for the following reactions at 179 °C. (a) PCl3(g) + Cl2(g)<--> PCl5(g) K = (b) 2 PCl5(g)<-->2 PCl3(g) + 2 Cl2(g) K =
The equilibrium constant, K, for the following reaction is 2.35×10-2 at 517 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 11.3 L container at 517 K contains 0.269 M PCl5, 7.94×10-2 M PCl3 and 7.94×10-2 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 5.27 L? [PCl5] = M [PCl3] = M [Cl2] =...