Question

Two tough chemistry questions?

In the following reaction, at equilibrium, there were 0.30 moles of N2(g), 2.0 moles of H2(g) and some amount of NH3(g) in a sealed 5.0 L flask. The KC for the reaction at this temperature is 25. If the reaction began with 1.0 moles of NH3(g), how much H2(g) did the reaction start with.
3 H2(g) + N2(g) >< 2 NH3(g)

Ammonium carbonate can decompose into ice, carbon dioxide and ammonia gas. When 3 moles of ammonium carbonate is placed in an empty, sealed 2.0 L flask and is allowed to reach equilibrium at -250C , 10 % of it decomposes. What is the value of KP at this temperature?
(NH4)2(CO3)(s) >< 2 NH3(g) + CO2(g) + H2O(s)

Answer 1

`H_2 = 12.13` moles

`K_p = 0.143`

Explanation:

From the equilibrium constant and final reactant amounts we can calculate the equilibrium amount of ammonia (7.75 moles). Reversin the "ICE" table we now can determine the "Change" in ammonia (6.75 moles) and thus calculate the change in nitrogen (-3.38) and hydrogen (-10.13) from the balanced chemical reaction.

The initial amount of hydrogen is then the final amount plus the 'change' amount, or 12.13 moles.

`K_c = 25 = [NH_3]^2/([H_2]^3 xx [N_2]`

`[NH_3]^2 = 25 xx ([2]^3 xx [0.3]) = 60`

`[NH_3] = 7.75`

`" " H_2" " N_2" " NH_3`
Initial`" "12.13 " "3.68" "1.0`
Change`" " -10.13" " -3.38" "6.75`
Equilibrium`" " 2.0" "0.3" "7.75`
`"------------------------------------------"`

`K_p = 0.143`
#(NH_4)_2(CO_3)(s) >< 2NH_3(g) + CO_2(g) + H_2O(s) #" " (NH_4)_2(CO_3)" " 2NH_3" " CO_2(g)" " H_2O# Initial#" "3.0" "0" "0# Change#" " -0.30" " +0.60" "+0.30# Equilibrium#" " 2.70" +0.60" "+0.30# #K_c = [NH_3]^2 xx [CO_2]/[(NH_4)_2(CO_3)]^2# #K_c = (0.6)^2 xx 0.3/2.7 = 0.040#

#K_p = K_c xx (R xx T)^(Delta n) Delta n = (number of moles of gaseous products - number of moles of gaseous reactants) #K_p = K_c xx (0.0821 xx 23)^2 = 0.040 xx 3.57 = 0.143#