An equilibrium mixture at 300 K contains ` N_2O_4 and NO_2` at 0.28 and 1.1 atmospheres respectively . If the volume of the container is doubled , calculate the new equilibrium pressures of the gases .

Answer

` p_(NO_2) = 0.605 " atm " , rho _(N_2O_4) = 0.085 ` atm

Explanation

The given equilibrium is
` " "N_2O_4(g) hArr 2NO_2 (g)`
Given that ` rho (N_2 O_4) = 0.28 " atm and " rho _(NO_2) = 1.1 " atm `
` therefore " " K_p = (rho_(NO_2) ^(2))/( rho _(N_2O_4) )= ( ( 1.1)^(2))/(0.28) = 4.32 " atm " `
When the volume of the container is doubled , the total pressure will reduce to half.
Total pressure of the system before doubling the volume
`" " = 0.28 + 1.1 = 1.38 ` atm
Pressure after doubling the volume of the container
` " " = (1.38)/(2)= 0.69 ` atm .
Suppose , under the changed circumstances , the degree of dissociation of `N_2O_4 " is " alpha ` , Therefore
` {:( ,, N_2O (g) , hArr, 2NO_2(g) ) , ( " Initial moles " , , 1, , 0 ),(" Moles at equilibrium ",, 1-alpha , , 2alpha ):}`
Hence ` " " rho N_2O_4 =(1-alpha)/(1+alpha) xx -0.69 `
and ` " " rho_(NO_2) = (2alpha )/(1+alpha) xx 0.69`
` because " " K_p = (rho _(NO_2) ^(2))/(rho_(N_2O_4))`
we have,
` 4.32 = (((2alpha)/(1+alpha ) xx 0.69 )^(2))/((1-alpha)/(1+alpha ) xx 0.69 )= (4alpha^(2) xx0.69)/(1-alpha ^(2))`
` or , " " (alpha^(2))/(1-alpha^(2)) = (4.32)/(4 xx 0.69) = 1.565`
` or , 1.565 (1-alpha^(2)) = alpha ^(2) `
` or , " " 2.565 alpha ^(2) = 1.565 `
` " " alpha = 0.781 `
Hence ,
` " " rho_(N_2O_4) = (1- 0.781 )/(1+0.781 ) xx 0.69 = 0.085 " atm" `
and ` " " rho _(NO_2) = (2xx 0.781 )/( 1+0.781 ) xx 0.69 = 0.605 ` atm.