For the dissociation of hydrogen cyanide in aqueous solution, i.e. #HC-=N(aq) + H_2O(l) rightleftharpoons H_3O^+ + ""^(-):C-=N#... which scenario will change the equilibrium constant?

#"a. adding more hydrocyanide;"#
#"b. raising the temperature;"#
#"c. adding more solvent;"#
#"d. adding another acid."#

1 Answer
Dec 28, 2015

Answer:

Only (b) will (possibly) change the equilibrium constant.

Explanation:

Hydrocyanic (prussic) acid undergoes the acid-base reaction as follows:

#HC-=N + H_2O rightleftharpoons ""^(-)C-=N + H_3O^+#

As with any equilibrium reaction, we can write the acid-base dissociation expression, #K_a# #=# #([""^(-)C-=N][H_3O^+])/[H_2O]#, which simplifies to #K_a# #=# #[""^(-)C-=N][H_3O^+]#.

Now I don't know the value of #K_a#; I am too lazy to look it up (you shouldn't be!); #K_a# will be reasonably small. #K_a# is quoted for conditions of standard temperature and pressure. I suspect that raising the temperature will increase #K_a# somewhat. Why? Because the above is a bond breaking reaction, and raising the temperature should allow greater bond cleavage. This question is also typically asked of the autoprotolysis of water; #pK_a# SHOULD be higher at elevated temperatures, #># #298K#, and indeed it is.

I take it that this is a 1st year problem?