Please help me Why are calculations of how much product is formed in the reactio

Question

Please help me Why are calculations of how much product is formed in the reaction X+Y equivalent Z often simpler when there is no Z initially present and the value of K is very small(strict inequality 10^-6)? Could the quadratic equation to be used to solve for the equilibrium concentration of NO_2 in the following reaction? 2NO(g)+O_2(g) equivalent 2NO_2(g) Consider this reaction: PCl_5(g) equivalent PCl_3(g)+Cl_2(g) K_p= 23.6 at 500 K Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are P_PCl_5= 0.560 atm and P_PCl_3= 0.500 atm. If more chlorine is added after the equilibrium is reached is reached, how will the concentrations of PCl_5 and PCl_3 change? Enough NO_2 gas is injected into a cylindrical vessel to produce a partial pressure, P_NO_2, of 0.900 atm at 298 K. Calculate the equilibrium partial pressures of NO_2 and N_2O_4, given 2NO_2(g) equivalent N_2O_4(g) K_P= 4 at 298 K

Explanation / Answer

Equilibrium constant K is defined as

K= [Z] /{[X][Y]}

If the initial concentration are M1 and M2 and equilibrium concentrations are M1-x and M2-x and Z formed is x

K= x/{M1-x)*(M2-x)

since K is small, x is small and M1-x can be approximated to M1 and M2 can also be approximated to M2. Hence

K= x/ M1.M2

x= K*M1M2. This Makes determination of x easier.

b)

Given initial pressures are PPcl5= 0.56 atm PPCl2= 0.5 atm

Let x =drop in pressure of PCl5 for establising equilibrium

At Equilibrium PPCl5= 0.56-x and PPCl3= 0.5+x and Pcl2= x

Kp =PPCl3* PPCl2/ PPCl5

Kp = x*(0.5+x)/ (0.56-x)= 23.6

This equation can easily be solved by using solver of excel, x= 0.5364

At Equilibrium PPCl5= 0.56-0.5364 =0.0236 atm and PPCl3= 0.5+0.5364=1.0364 atm and PCl2=0.5364 atm

if more chlorine is injected, more moles on the product side and as per Lechatlier principle, more PPCl5 will have to be there to keep the equilibrium constant same.

Initial partial pressures :   NO2 = 0.9 atm and   N2O4=0

Let 2x= drop in partial pressure of NO2 to establsih equilibrium

Equilibrium                         NO2= 0.9-2x   and N2O4=x

Kp =[PN2O4]/ {PNO2}2 = x/(0.9-2x)2= 4. This can be solved by solver which gives

The value of x = 0.3107

PN2O4= 0.3107 atm and PNO2= 0.9-2*0.3107=0.2786 atm

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