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# PCl3(g)+Cl2(g)<—>PCl5(g) To a empty 15.0 L cylinder, 0.500 moles of gaseous PCl5 are added and allowed to reach equilibrium. ?

The concentration of the PCl3 is found to be 0.0220 M. Assume a temperature of 375 K and all concentrations are molar concentrations.

a) How many moles of PCl3 remain at equilibrium?

b) Write the equilibrium constant expression for the above reaction.

c) Determine the value of Kc.

d) Determine the value of Kp for this same system at the same temperature

e) How would the value of Kp be effected by increasing the temperature of the system at equilibrium for this exothermic reaction.

### 2 Answers

- davidLv 71 month agoFavorite Answer
PCl3 + Cl2 <--> PCl5

initial [PCl5] = 0.50/15.0L = 0.0333333 M

equilib [PCl5] = 0.0220M

change = 0.0113 M used t reach equilib.

equalib [PCl3] = equiib [Cl2] = 0.0113 M

a. moles PCl3 at eq = 0.0113 M X 15.0L = 0.1695M >>> round for sig figs

b. Kc = [PCl5] / ([PCl3][Cl2]

c. Kc = 0.03333 / (0.1695 X 0.1695) === you can use a calculator for this

d. There is a formula for Kp when you know Kc === you either divide by R^(delta n) or multiply ... look it up == you can do this.

e. Kp would decrease if Temp is increased

- jacob sLv 71 month ago
a) How many moles of PCL5 remains in equilibrium?

Molarity PCl5 = mole/L -> 0.0220M = mole/15-> 0.330 mol PCL5

Since is a 1:1 ratio, that means there are 0.170 mol PCL5 remaining

b) What is the equilibrium constant for the reaction above?

Ka = [PCl5]/[PCL3][CL2]

c) Determine the value of K

Ka = (0.170/15)/[0.0220][0.0220] = 23.4

d . Kp =0.75 Answer

e. Kp would decrease . Increasing the temperature of an exothermic reaction would shift the reaction to the left , increasing the denominator and decreasing the numerator.