Question

Use the given information to determine the enthalpy of the reaction

Answer 1

The question is incomplete, here is the complete question.

Consider these reactions, where M represents a generic metal.

1. $2M(s)+6HCl(aq)\rightarrow 2MCl_3(aq)+3H_2(g)$;   ΔH = -783 kJ

2. $HCl(g)\rightarrow HCl(aq)$;   ΔH = -74.8 kJ

3. $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$;   ΔH = -1845 kJ

4. $MCl_3(s)\rightarrow MCl_3(aq)$;  ΔH = -195 kJ

Use the information above to determine the enthalpy of the following reaction.

$2M(s)+3Cl_2(g)\rightarrow 2MCl_3(s)$;    ΔH=?

Answer : The enthalpy for the reaction is, -6376.8 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Now we have to determine the enthalpy of the reaction.

Multiplying reaction 2 by 6, reaction 3 by 3 and reverse reaction 4 by 2 then adding all the reaction, we get:

1. $2M(s)+6HCl(aq)\rightarrow 2MCl_3(aq)+3H_2(g)$;   ΔH = -783 kJ

2. $6HCl(g)\rightarrow 6HCl(aq)$;    ΔH = 6 × (-74.8 kJ) = -448.8 kJ

3. $3H_2(g)+3Cl_2(g)\rightarrow 6HCl(g)$;   ΔH = 3 × (-1845 kJ) = -5535 kJ

4. $2MCl_3(aq)\rightarrow 2MCl_3(s)$;   ΔH = 2 × 195 kJ  = 390 kJ

The expression for enthalpy for the reaction is,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-783)+(-448.8)+(-5535)+(390)$

$\Delta H=-6376.8kJ$

Therefore, the enthalpy for the reaction is, -6376.8 kJ