Chemistry Forum

Can somoone plesa see this exercise?

For the reaction PCl₃+Cl₂-PCl₅, at a certain temperature k=4. Calculate the speed of this reaction when the concentration of PCl₃ is 3 mol/l and the concentration of Cl₂ is 2 mol/l. What is the speed of this reaction at this temperature when there are 1.5 mol of Cl₂ left in the container?

K = cPCl₅/(cPCl₃*cCl₂)

4 = x/((3-x)*(2-x)) solve for x.

The other reaction is
If 1,5 mol/ l Cl₂ is left then 0,5 mol/l has reacted. What means the same amount of PCl₃ has reacted and also the same amount PCl₅ was obtained.

K= 0,5/((3-0,5)*(2-0,5))

Okat then a) 24 mol/l*s
b) 15 mol/l*s is this right?

Yes I would think so