1.

For the decomposition reaction NH_2COONH_4 (s) hArr 2NH_3(g) +CO_2(g) The K_p=2.9 xx 10^(-5) atm^3. The total pressure of gases at equilibrium when 1 mole of NH_2COONH_4 (s) was taken to start with would be

Answer»

0.0194 atm
0.0388 atm
0.0582 atm
0.0766 atm

Solution :`{:(NH_2COOHN_4(s), HARR 2NH_3(G), +CO_2(g)),(1,2,1):}`
`K_p=2.9xx10^(-5)atm^3`
If P is the total pressure at equilibrium
`K_p=((2p)/3)^2(p/3)`
`THEREFORE P_3=(27xx2.9xx10^(10^(-5)))/4=1.9575`
`P=3sqrt(1.9575)=0.0582`


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