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The equilibrium pressure of NH4CN(s)⇌NH3(g)+HCN(g)  is 0.298 atm.If NH4CN(s)  is allowed to decompose in presence of NH3  at 0.25 atm,calculate partial pressure of HCN at equilibrium .If partial pressure of HCN at equilibrium is   y×10−2  atm ,then what is y

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a

3

b

5

c

7

d

9

answer is C.

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Detailed Solution

NH4CN⇌NH3+HCN                           0.15     0.15 Kp=0.15×0.15=0.0225   atm2 NH4CN⇌NH30.25+HCN0x                     0.25+x     x Kp=(0.25+x)x 0.0225=(0.25+x)x x=7×10−2
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The equilibrium pressure of NH4CN(s)⇌NH3(g)+HCN(g)  is 0.298 atm.If NH4CN(s)  is allowed to decompose in presence of NH3  at 0.25 atm,calculate partial pressure of HCN at equilibrium .If partial pressure of HCN at equilibrium is   y×10−2  atm ,then what is y