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 :

An arbitrary combound P_2Q decomposes according to reaction : 2P_2Q(g

1.	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 :
Answer» mole of `.^nP_2Q=.^nQ_2` MOLES of `.^nP_2=(8//3)` DEGREE of dissociation is `alpha =(2//3)` total number of moles of products (`P_2` and `Q_2`) at equilibrium is 4 Solution :`{:(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`