If equilibrium pressure is \(6\) atm for the above reaction; \(K_p\) will be:
\(NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2 (g)\)
1
\(32\)
2
\(27\)
3
\(\dfrac{4}{27}\)
4
\(\dfrac{1}{27}\)
\(NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2 (g)\)