questions in the picture below (show your works) A flask is charged with 0.750 a

Question

questions in the picture below (show your works)

A flask is charged with 0.750 atm of dinitrogen tetroxide gas and 0.600 atm of nitrogen dioxide gas at 25degreeC. After equilibrium is reached between these two gases, the partial pressure of dinitrogen tetroxide is 0.118 atm (a) Write balanced equation for reaction (NO_2 is reactant) and complete following ICE Table Calculate K_p for the reaction at 25degree C If an additional 0.150 atm of N_2O_4 were added to the above equilibrium mixture, what would be resulting partial pressures of both components at the new equilibrium condition ?

Explanation / Answer

N2O4-ßà 2NO2

Partial pressure of N2O4= 0.75 atm and that of NO2= 0.6 atm

Let x= drop in partial pressure of N2O4 to reach equilibrium

At equilibrium Partial pressure of N2O4= 0.75-x and partial pressure of NO2= 0.6+2x

0.75-x =0.118 and x =0.75-0.118=0.632 atm

So at equilibrium

Partial pressures : N2O4= 0.118 atm and NO2= 0.6+2*0.118=0.836 atm

Kp = [PNO2]2/ [PN2O4] = 0.836*0.836/0.118=5.9228

When 0.150atm of N2O4 is added, partial pressure of N2O4 = 0.118+0.150 =0.268 atm

Let x= drop in partial pressure of N2O4 to react new equilibrium

5.9228= (0.836+2x)2/( 0.268-x) when solved x= 0.092 atm

So at new eq. N2O4= 0.268-0.092= 0.176 atm and NO2= 0.836+2*0.092=1.02 atm

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