1.

Assume that the decomposition of HNO_(3) can be represented by the following equation 4NHO_(3)(g)hArr4NO_(2)(g)+2H_(2)O(g)+O_(2)(g) and the reaction approaches equilibrium partial pressure of HNO_(3) is 2 atm. Calculate K_(e) in ((mol)/L)^(3) at 400K : (Use : R = 0.08 atm-L/mol-K)

Answer»

4
8
16
32

Solution :`P_("total")=P_(HNO_(3))+P_(NO_(2))+P_(H_(2)O)+P_(O_(2))`
`because P_(NO_(2))=4P_(O_(2))and P_(H_(2)O)=2P_(O_(2))`
`THEREFORE P_("total")=P_(HNO_(3))+7PO_(2)`
`rArr30-2=P_(O_(2))xx7`
`rArr28/7=4`
`K_(p)=(P_(NO_(2))^(4)cdotP_(H_(2)O)cdotPO_(2))/P_(HNO_(3))^(4)`
`=((4xx4)^(4)xx(2xx4)^(2)xx4)/(2^(4))=2^(20)`
`K_(p)=K_(e)(RT)^(Deltan_(g))=K_(c)(0.08xx400)^(3)`
`rArrK_(c)=2^(20)/(32)^(3)=32`


Discussion

No Comment Found

Related InterviewSolutions