determining enthalpy of the reaction

Question

Consider these reactions, where M represents a generic metal. 1. 2M(s) + 6HCl(aq) --- 2MCl3(aq) + 3H2(g) delta H1= -941.0 kJ 2. HCl(g) --- HCl (aq) delta H2= -74.8 kJ 3. H2(g) +Cl2(g) --- 2HCl(g) delta H3= -1845.0 kJ 4. MCl3 (s) --- MCl3(aq) delta H4= -422.0 kJ Use the given information to determine the enthalpy of the reaction 2M(s) + 3Cl2(g) --- 2MCl3 (s)

Answer 1

For more info on how to approach these types of questions read up on Hess' Law. Use the given equations and rearrange them in order to obtain the target equation when they are added together. If you reverse an equation, you must change the sign of ∆H. If you multiply the equation by a value, you must multiply ∆H by that same value.

2M(s) + 3Cl2(g) ==> 2MCl3(s) .. TARGET EQUATION

Given:

(1) 2M(s) + 6HCl(aq) ==> 2MCl3(aq) + 3H2(g) ... ∆H = -941.0 kJ

(2) HCl(g) ==> HCl(aq) ... ∆H = -74.8 kJ

(3) H2(g) + Cl2(g) ==> 2HCl(g) ... ∆H = -1845.0 kJ

(4) MCl3(s) ==> MCl3(aq) ... ∆H = -422.0 kJ

Procedure:

copy (1): 2M(s) + 6HCl(aq) ==> 2MCl3(aq) + 3H2(g) ... ∆H = -941.0 kJ

reverse (4) and multiply by 2: 2MCl3(aq) ==> 2MCl3(s) ... ∆H = +844.0 kJ

multiply (3) by 3: 3H2(g) + 3Cl2(g) ==> 6HCl(g) ... ∆H = -5535 kJ

multiply (2) by 6: 6HCl(g) ==> 6HCl(aq) ... ∆H = -448.8 kJ

add them all together and then combine/cancel like terms:

2M(s) + 6HCl(aq) + 2MCl3(aq) + 3H2(g) + 3Cl2(g) + 6HCl(g) => 2MCl3(aq) + 3H2(g) + 2MCl3(s)+ 6HCl(g) + 6HCl(aq)

This leaves us with the TARGET EQUATION: 2M(s) + 3Cl2(g) ==> 2MCl3(s)

To get the enthalpy of the reaction, add up all the ∆H values:

∆H = -941 + 844 - 5535 -449 = 6081 kJ

(be sure to check all of the math)