Question #e4ee3

1 Answer
Jun 12, 2015

Answer:

The change in enthalpy for your reaction will be #DeltaH = "-78.5 kJ"#.

Explanation:

I'll start by saying that the equations you wrote make no sense. Not only did you mistype compounds, but you wrote the same incorrect reaction twice.

Your target reaction is

#HCl + NaNO_2 -> HNO_2 + NaCl#, #" "DeltaH_"rxn"=?#

The four reactions that you're supposed to use look like this:

#color(blue)((1))# #2NaCl + H_2O -> 2HCl + Na_2O#, #" "DeltaH_1 = "+507 kJ"#

#color(blue)((2))# #NO + NO_2 + Na_2O -> 2NaNO_2#, #" "DeltaH_2 = "-427 kJ"#

#color(blue)((3))# #NO + NO_2 -> N_2O + O_2#, #" "DeltaH_3 = "-43 kJ"#

#color(blue)((4))# #2HNO_2 -> N_2O + O_2 + H_2O#, #" "DeltaH_4 = "+34 kJ"#

Take a look at the four reactions you have. Notice that the target equation has sodium chloride on the products's side and hydrochloric acid on the reactants' side.

Reaction #color(blue)((1))# has them the other way around. Not only that, but this reaction has 2 moles of each, instead of 1 mole of each. Don't worry about that right now, just flip reaction #color(blue)((1))#. This will become

#color(blue)((1^'))# #2HCl + Na_2O -> 2NaCl + H_2O#,

#DeltaH_1^' = (-DeltaH_1) = "-507.31 kJ" = "-507 kJ"#

Now focus on sodium nitrite. Once again, the target reaction has it on the reactants' side, but reaction #color(blue)((2))# has it on the products's side.

Flip reaction #color(blue)((2))# to get

#color(blue)((2^'))# #2NaNO_2 -> NO + NO_2 + Na_2O#

#DeltaH_2^'= (-DeltaH_2) = "+ 427.14 kJ" = "+427 kJ"#

Finally, focus on nitrous acid. You need it on the products' side, but have it on the reactants' side in reaction #color(blue)((4))#. Once again, flip reaction #color(blue)((4))# to get

#color(blue)((4^'))# #N_2O + O_2 + H_2O -> 2HNO_2#

#DeltaH_4^' = (-DeltaH_4) = "-34.25 kJ" = "-34 kJ"#

You have everything that you need for the target reaction, so leave reaction #color(blue)((3))# unchanged.

Add all these reactions and eliminate compounds that are on both sides of the general reaction to get

#2HCl + cancel(Na_2O) + 2NaNO_2 + cancel(NO) + cancel(NO_2) + cancel(Na_2O) + cancel(O_2) + cancel(H_2O) = 2NaCl + cancel(H_2O) + cancel(NO) + cancel(NO_2) + cancel(Na_2O) + cancel(N_2O) + cancel(O_2) + 2HNO_2#

This is reduced to the form

#2HCl + 2NaNO_2 -> 2NaCl + 2HNO_2#

#DeltaH_"rxn" = DeltaH_1^' + DeltaH_2^' + DeltaH_3 + DeltaH_4^'#

#DeltaH_"rxn" = "-507" + 427 + ("-43") + ("-34 kJ")#

#DeltaH_"rxn" = "-157 kJ"#

All you have to do now is divide this reaction by 2 to get the final form

#HCl + NaNO_2 -> NaCl + HNO_2#

#DeltaH_"rxn" = ("-157 kJ"/2) = color(green)("-78.5 kJ")#