Question

Consider these reactions, where M represents a generic metal. 2 M ( s ) + 6 HCl ( aq ) ⟶ 2 MCl 3 ( aq ) + 3 H 2 ( g ) Δ H 1 = − 760.0 k J HCl ( g ) ⟶ HCl ( aq ) Δ H 2 = − 74.8 k J H 2 ( g ) + Cl 2 ( g ) ⟶ 2 HCl ( g ) Δ H 3 = − 1845.0 k J MCl 3 ( s ) ⟶ MCl 3 ( aq ) Δ H 4 = − 267.0 k J Use the given information to determine the enthalpy of the reaction 2 M ( s ) + 3 Cl 2 ( g ) ⟶ 2 MCl 3 ( s )

Answer 1

Answer : The enthalpy of the reaction is, -6209.8 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.

The Main reaction is:

`2M(s)+3Cl_2(g)\rightarrow 2MCl_3(s)`
`\Delta H=?`

The intermediate balanced chemical reaction will be,

(1)
`2M(s)+6HCl(aq)\rightarrow 2MCl_3(aq)+3H_2(g)`
`\Delta H_1=-760.0kJ`

(2)
`HCl(g)\rightarrow HCl(aq)`
`\Delta H_2=-74.8kJ`

(3)
`H_2(g)+Cl_2(g)\rightarrow 2HCl(g)`
`\Delta H_3=-1845.0kJ`

(4)
`MCl_3(s)\rightarrow MCl_3(aq)`
`\Delta H_4=-267.0kJ`

We are multiplying equation 3 by 3, reverse reaction of 4 by 2 and reaction 2 by 6 and then adding all the equations, we get :

(1)
`2M(s)+6HCl(aq)\rightarrow 2MCl_3(aq)+3H_2(g)`
`\Delta H_1=-760.0kJ`

(2)
`6HCl(g)\rightarrow 6HCl(aq)`
`\Delta H_2=6* (-74.8kJ)=-448.8kJ`

(3)
`3H_2(g)+3Cl_2(g)\rightarrow 6HCl(g)`
`\Delta H_3=3* (-1845.0kJ)=-5535.0kJ`

(4)
`2MCl_3(aq)\rightarrow 2MCl_3(s)`
`\Delta H_4=2* 267.0kJ=534.0kJ`

The expression for enthalpy of the reaction is,

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

`\Delta H=(-760.0kJ)+(-448.8kJ)+(-5535.0kJ)+(534.0kJ)`

`\Delta H=-6209.8kJ`

Therefore, the enthalpy of the reaction is, -6209.8 kJ