Q.

For the decomposition reaction NH2COONH4(s)2NH3(g)+CO2(g),  the Kp=2.9×105atm3. The total pressure of gases at equilibrium when 1.0mol of  NH2COONH4(s) was taken to start with would be

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a

0.0194 atm

b

0.0776 atm

c

0.0388 atm

d

0.0582 atm

answer is C.

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Detailed Solution

NH2COONH4(s)1.0molx2NH3(g)2p+CO2(g)p

Kp=pNH32pCO2 i.e. 2.9×105atm3=(2p)2p

This gives p=2.9×105atm41/3=0.0194atm and  ptotal =2p+p=3p=3×0.0194atm=0.0582atm

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