Given the hypothetical reaction:2A(8) + nB(g) 3Ce) K-0.0105and K 0.45

1.	Given the hypothetical reaction:2A(8) + nB(g) 3Ce) K-0.0105and K 0.45 at 250° C. What is the value of coefficientn' ?
Answer» Kp = Kc(RT)^(∆ng) 0.0105 = 0.45(0.082523)^(3-n)0.0105/0.45= (43)^(3-n)=> 0.023 = (43)^(3-n) log(0.023) = (3-n)log43=> -1.68 = 1.63(3-n)=> 1 = 3-n=> n = 3-1 = 2 bit how did you get 3-n it Must be 1-n no.. we always look for gasses... here A is (s) solid.. so it will not count. that's why i wrote ∆ng ~ ∆n gases . ohh thanks a lot

Given the hypothetical reaction:2A(8) + nB(g) 3Ce) K-0.0105and K 0.45 at 250° C. What is the value of coefficientn' ?

Answer»

Kp = Kc(RT)^(∆ng)

0.0105 = 0.45*(0.082*523)^(3-n)0.0105/0.45= (43)^(3-n)=> 0.023 = (43)^(3-n) log(0.023) = (3-n)log43=> -1.68 = 1.63(3-n)=> 1 = 3-n=> n = 3-1 = 2

bit how did you get 3-n

it Must be 1-n

no.. we always look for gasses... here A is (s) solid.. so it will not count.

that's why i wrote ∆ng ~ ∆n gases .

ohh thanks a lot