Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
Explanation :
For the given chemical reaction:
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex] ........(1)
[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]
Initial: 0 0
At eqm: 2x x
As we are given that:
Total pressure of gas at equilibrium = 0.116 atm
2x + x = 0.116 atm
3x = 0.116 atm
x = 0.0387 atm
Putting values in expression 1, we get:
[tex]K_p=(2x)^2\times (x)[/tex]
[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]
[tex]K_p=2.32\times 10^{-4}[/tex]
Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]
