ALL CIE CHEMISTRY DOUBTS HERE !!

[moon]

43 - 7 = 36

Therefore 95 - 36 = 59 pm. IDK

sorry Ancestor I didn't get it.

[Arthur Bon Zavi]

sorry Ancestor I didn't get it.

Do you know the answer?

[moon]

Do you know the answer?

yes, the answer is 78 but how did we find out this value

[Arthur Bon Zavi]

yes, the answer is 78 but how did we find out this value

Sorry, that was wrong method. 

[Dania]

Can someone please help me on October November 2006 Paper 2 no. 5 (e)?

[SpongeBob]

I guess I'm late =(

[Deadly_king]

I guess I'm late =(

I don't think so.

Do you mind to help Dania for her chem number, please?

Am pretty busy for now koz i've got exams tomorrow

[ashish]

Can someone please help me on October November 2006 Paper 2 no. 5 (e)?

her is your answer check out the attachment

[Uchia]

when some phosphorus penta chloride was heated to a temperature T in a container 78% OF IT decomposed and the pressure in the container was 2.2


PCl5(g)  backward  and forward reaction      PCL3(g) + CL2 (g)


could you please help me to solve this question and thanks

[Uchia]

calculate Kp

[Master_Key]

Kp = (p[Cl₂] * p[PCl₃]) / p[PCl₅]

[Arthur Bon Zavi]

when some phosphorus penta chloride was heated to a temperature T in a container 78% OF IT decomposed and the pressure in the container was 2.2


PCl5(g)  backward  and forward reaction      PCL3(g) + CL2 (g)


could you please help me to solve this question and thanks

PCl₅ <--------------> PCl_3(g)  +  Cl_2(g)

1 mole of phosphorous-penta-chloride was heated, i.e 100 %, 78 % decomposed ! i.e 78 % of 1 mole..

Therefore : 78 / 100 = 0.78 moles

                              [PCl₃] X [Cl₂]
 K_p =      ----------------------------------------
                                    [PCl₅]

Total : 0.78 + 0.78 + 0.78 = 2.34

Fraction of PCl5 =  0.78        1
                            ------- =  --- (Same is for PCl₃ and Cl₂).
                             2.34        3

2.2 MPa X 1000 = 2200 KPa

(1/3) (2200) = 733.3 ----------------------> Partial pressure for the three.

Therefore K_p = (733.3)² / 733.3 = 733.3 KPa

Is this the answer ?

[Uchia]

mm no

anyways thnxs for your help I found the answer

Initial          PCl5                           PCl3           Cl2
                    1                                0               0
Change
                 1-0.78=0.22                0.78        0.78     

Total mole 0.22 + 0.78 + 0.78 = 1.78

(0.22/1.78) * 2.2
(0.78/1.78) * 2.2
(0.78/1.78) * 2.2

then continue

see the problem was that       for PCl5 you take the amount left the one that was unreacted this is done by subtraction because at equilibrium you should use the amount of both the reactants and the products in the experimen and using the amount that decomposed could not be used because it already changed to products

[Arthur Bon Zavi]

mm no

anyways thnxs for your help I found the answer

Initial          PCl5                           PCl3           Cl2
                    1                                0               0
Change
                 1-0.78=0.22                0.78        0.78     

Total mole 0.22 + 0.78 + 0.78 = 1.78

(0.22/1.78) * 2.2
(0.78/1.78) * 2.2
(0.78/1.78) * 2.2

then continue

see the problem was that       for PCl5 you take the amount left the one that was unreacted this is done by subtraction because at equilibrium you should use the amount of both the reactants and the products in the experimen and using the amount that decomposed could not be used because it already changed to products

Oh! Yeah. Sorry. Just near.

But was that 2.2 MPa Or KPa ?

[Uchia]

kpa maybe