Question

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s) ⇀↽ 2 NH3(g) + CO2(g) The total pressure in the closed container under these condition is found to be 0.843 atm. Calculate a value for the equilibrium constant, Kp. A) 0.00701 B) 0.0888 C) 0.222 D) 0.599

Answer 1

Value of equilibrium constant is 0.0888

Step-by-step explanation:

Both
`NH_(3)` and
`CO_(2)` are gaseous. Hence equilibrium constant depends upon partial pressures of
`NH_(3)` and
`CO_(2)`.

Initially no
`NH_(3)` and
`CO_(2)` were present.

Hence mole fraction of
`NH_(3)` and
`CO_(2)` at equilibrium can be calculated from coefficient of
`NH_(3)` and
`CO_(2)` in balanced equation.

Mole fraction of
`NH_(3)` = (number of moles of
`NH_(3)`)/(total number of moles of
`NH_(3)` and
`CO_(2)`) =
`(2moles)/((2+1)moles)=(2)/(3)`

Mole fraction of
`CO_(2)` = (number of moles of
`CO_(2)`)/(total number of moles of
`NH_(3)` and
`CO_(2)`) =
`(1moles)/((2+1)moles)=(1)/(3)`

Let's assume both
`CO_(2)` and
`NH_(3)` behaves ideally.

Therefore partial pressure of
`NH_(3)`,
`P_{NH_(3)}= x_{NH_(3)}.P_(total)` and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So,
`P_{NH_(3)}=(2)/(3)* 0.843atm=0.562atm`

`P_{CO_(2)}=(1)/(3)* 0.843atm=0.281atm`

So,
`K_(p)=P_{NH_(3)}^(2).P_{CO_(2)}=(0.562)^(2)* 0.281=0.0888`