Question

A chemist measures the enthalpy change ΔH during the following reaction:

2HNO3(l)+Mg(s)→Mg(NO3)2(s)+H2(g) =ΔH−443.kJ

Use this information to complete the table below. Round each of your answers to the nearest/kJmol.

Reaction ΔH

Mg(NO3)2(s)+H2(g) → 2HNO3(l)Mg(s) =kJ

6HNO3(l)+3Mg(s) →3Mg(NO3)2(s)3H2(g) =kJ

1/2Mg(NO3)2(s)+1/2H2(g) → HNO3(l)+1/2Mg(s) =kJ

Answer 1

Consider the following reaction as follows 2HNO, ()+ Mg (s)- Mg(NO,), (s)+H,(g) The reaction is exothermic A 443kJ Now consider the reaction as follows Mg (NO,), (s)+H2 (g) 2HNO, ()+ Mg(s) This reaction is reverse of the above reaction (1) When the reaction is reversed, the heat change of the reaction changes signs. That is, the new reaction proceeds with absorption of energy and is endothermic as follows Mg (NO3 )2 (s)-H2 (g)-2HNO3 (1)-Mg (s) ΔΗ-443kJ Now consider the reaction as follows 6HNO3 (1)+3Mg (s) . 3Me (NO.).(s)-3H2 (g) This reaction can be better written as follows: 3 2HNO, ()+ Mg (s)-Mg (NO, ) (s)+H , (8)) That is, 3*multiple of equation (1) When the reaction is multiplied its heat change also get multiplied with the same number Therefore, the heat change of the above reaction will be 3(-443kJ)1329kJ. The reaction is written as follows: 6HNo, (l) +3Mg (s) → 3Mg (NO3 )2 (s) +3H2 (g) AH1329 kJ Now consider the reaction as follows 1/2Mg (NO,)s+/2H (g)HNO,()+1/2Mg(s) The reaction is reverse and 1/2 multiple of equation (1) When the reaction is multiplied its heat change also get multiplied with same number Therefore, the heat change of the reaction will be-443kJ)--221.5kJ The reverse reaction changes its sign from negative to positive The reaction is written as follows: 1/2Mg (NO3)2(s)+1/2H2 (g) →HNO3 (1)+1/2Mg (s)