An arbitrary combound `P_2Q` decomposes according to reaction : `2P_2Q(g)hArr 2P_2(g)+Q_2(g)` If one starts the decomposition reaction with 4 moles of
An arbitrary combound `P_2Q` decomposes according to reaction :
`2P_2Q(g)hArr 2P_2(g)+Q_2(g)`
If one starts the decomposition reaction with 4 moles of `P_2Q` and value of equilibrium constant `K_P` is numerically equal to total pressure at equilibrium.Then which option is/are correct at equilibrium :
A. mole of `.^nP_2Q=.^nQ_2`
B. moles of `.^nP_2=(8//3)`
C. degree of dissociation is `alpha =(2//3)`
D. total number of moles of products (`P_2` and `Q_2`) at equilibrium is 4
1 Answers
Correct Answer - A,B,C,D
`{:(2P_2Q,hArr,2P_2+,Q_2),(t=0,4,0,0),(t=eq,4(1-alpha),4alpha,2alpha):}`
`K_p=(((4alpha)/(4+2alpha)xxP)^2((2alpha)/(4+2alpha)xxP))/(((4(1-alpha))/(4+2alpha)xxP)^2)=(2alpha^3)/((1-alpha)^2(4+2alpha))xxP`
But `K_p=P` (given) `:. 2alpha^3=(1-alpha)^2(4+2alpha)`
`:. -6alpha+4=0 " " alpha=(2//3)`
Total moles at eq `=4+2alpha=(16//3)`
`n_(P_2Q)=n_(Q_2)=(4//3),n_(P_2)=(8//3)`
`:.` Total number of moles of products `=8/3+4/3=12/3=4`