NH4HS=NH3+H2S the equilibrium pressure at 25 degree Celsius is 0.8 atmosphere then kp for the reaction is ?

#K_P^@ = 0.16#

Well, if we write out the ICE table, it may become clearer...

#"NH"_4"HS"(s) rightleftharpoons "NH"_3(g) + "H"_2"S"(g)#

#"I"" "-" "" "" "" "" "0" "" "" "" "0#
#"C"" "-" "" "" "" "+P_i" "" "+P_i#
#"E"" "-" "" "" "" "" "P_i" "" "" "P_i#

One thing that was NOT mentioned was that #"NH"_4"HS"# is a solid at room temperature... hence, it shows up in #K_P# as #1#.

Assuming the gases are ideal, they have the same partial pressure at equilibrium (with coefficients of #1#).

Hence, the total pressure is:

#P = 2P_i = "0.8 atm"#

and #K_P^@# (i.e. the #K_P# at standard temperature) in terms of the partial pressures is:

#K_P^@ = ((P_(NH_3)//P^@)(P_(H_2S)//P^@))/((["NH"_4"HS"]//c_(NH_4HS)^@))#

The "concentration" of #"NH"_4"HS"# is the mols of it contained in #"1 L"# of the sample, which has a density of #"1.17 g/mL"#:

#"1.17 g"/"mL" xx "1000 mL"/"1 L" xx "1 mol"/"51.112 g" = "22.891 mol/L"#

The standard concentration #c^@# of #"NH"_4"HS"# is its "molar density" at #25^@ "C"# and #"1 atm"#, which is also #"22.891 mol/L"#. Hence, since standard pressure is chosen to be #"1 atm"# here,

#color(blue)(K_P^@) = (P_i//P^@)^2/(cancel(("22.891 mol/L")//("22.891 mol/L"))^(1))#

#= ((1/2 cdot 2P_i)/P^@)^2#

#= ((1/2 cdot 0.8 cancel"atm")/cancel"1 atm")^2#

#= color(blue)(0.16)#