Map sapling learning a) Calculate the standard heat of the neutralization reacti
ID: 853754 • Letter: M
Question
Map sapling learning a) Calculate the standard heat of the neutralization reaction between dilute hydrochloric acid and dilute potassium hydroxide, given that the heat of solution of potassium hydroxide is -54.01 kJ/mol, the heat of formation of potassium hydroxide is -426.84 kJ/mol, the heat of solution of potassium chloride is 18.426 kJ/mol, and the heat of formation of potassium chloride is -436.584 kJ/mol. Number kJ mol AH b) Calculate the heat of reaction for the reaction of gaseous HCl and solid KOH to form liquid water and solid KCI. Number AH kJ molExplanation / Answer
HCl (aq) + KOH (aq)----> KCl (aq) + H2O (l)... H = -57.1 kj/mole
hence enthalpy of neutralization of HCl with KOH is 57.1 kj enthalpy of neutralization of any strong acid (like HCl,HNO3,H2SO4) with a strong base (like LiOH,KOH,KOH) or vice versa is always the same i.e. 57.1 kj...this is because strong acids ,strong bases and salt that they form are all completely ionized in dilute aqueous solutions ...thus the reaction between any strong acid and strong base for example in the above case may be written as :
KOH (aq) + HCl(aq) -----> KCl (aq) + H2O (l)... H = -57.1 kj/mole
they will dissociate as :
K(+) (aq) + OH(-) (aq) + H(+) (aq) + Cl(-) (aq) ---> K(+) (aq) + Cl(-) (aq) + H2O (l)
common ions will cancel out..
H(+) (aq) + OH(-) (aq) ----> H2O (l)
thus neutralization is simply a reaction between H(+) ions given by acids and OH(-) ions given by base to form one mole of H2O.....since strong acid and strong base completely ionize in aqueous solution number of H(+) and OH(-) produced by 1 gram equivalent of strong acid and strong base is always the same ...hence enthalpy of neutralization between a strong acid and strong base is always constant...
ANSWER - H = -57.1 kj/mole