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An equilibrium mixture at 300 K containsN _(2) O_(4) and NO_(2) at 0*28 and 1*1 atmrespectively. If the volume of the container is doubled, calculate the new equilibrium pressures of two gases. |
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Answer» Solution :Step 1 . Calculate of `K_(p)` ` {:(,N_(2)O_(4),hArr,2 NO_(2)),("Equilibrium",0*28 "ATM",,1*1"atm"):}` `K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= (1*1 "atm")^(2)/(0*28 "atm")=4*32 "atm"` Step 2 . Calsulation of new equilibrium pressures . On doubling the volume , PRESSURE will decreaseto half . Hence, equilibrium will shift to the sideaccompanied by increasein the number of moles , i.e.,forward direction . This MEANS that pressure of `N_(2)O_(4) `will decreasewhile that of `NO_(2)`will increase . Suppose decreasein pressureof `N_(2)O_(4) = p` . Then ` {: (,N_(2)O_(4),hArr,2NO_(2)), ("Intial pressures ",0*28//2 "atm",,1*1//2 "atm"), ("New eqm. pressures ",(0*28/2-p)"atm",,(1*1/2 +2 p)"atm" ),(,=(0*14 -p)"atm",,=(0*55 +2p) "atm"):} ` . ` 0* 3025 + 4p^(2) + 2*2 p = 0* 6048 - 4* 32 p ` ` 4p^(2) + 6*52 p - 0*3023= 0 ` ` p = (-6* 52 pm sqrt(42 *51 + 4*84 ))/8 = 0*045 "atm" ` (minus value is neglected ) For quadratic equation ` ax^(2) + bx + c = 0 : x = (-b pm sqrt(b^(2) - 4ac))/ (2a)` ` :." New equilibrium pressures ")` ` p_(N_2O_(4)) = 0 *14 - 0* 045 = 0* 095 "atm" ` ` p_(NO_(2)) = 0* 55 + 2 xx 0* 045 = 0* 64 " atm"` ` |
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