Question

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241.8 kJ 2 ) X ( s ) + 2 Cl 2 ( g ) ⟶ XCl 4 ( s ) Δ H 2 = + 461.9 kJ 3 ) 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) ⟶ HCl ( g ) Δ H 3 = − 92.3 kJ 4 ) X ( s ) + O 2 ( g ) ⟶ XO 2 ( s ) Δ H 4 = − 789.1 kJ 5 ) H 2 O ( g ) ⟶ H 2 O ( l ) Δ H 5 = − 44.0 kJ what is the enthalpy, Δ H , for this reaction? XCl 4 ( s ) + 2 H 2 O ( l ) ⟶ XO 2 ( s ) + 4 HCl ( g )

Answer 1

Answer : The enthalpy of the given reaction will be, -1048.6 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.

The main reaction is:

`XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)`
`\Delta H=?`

The intermediate balanced chemical reactions are:

(1)
`H_2(g)+(1)/(2)O_2(g)\rightarrow H_2O(g)`
`\Delta H_1=-241.8kJ`

(2)
`X(s)+2Cl_2(g)\rightarrow XCl_4(s)`
`\Delta H_2=+461.9kJ`

(3)
`(1)/(2)H_2(g)+(1)/(2)Cl_2(g)\rightarrow HCl(g)`
`\Delta H_3=-92.3kJ`

(4)
`X(s)+O_2(g)\rightarrow XO_2(s)`
`\Delta H_4=-789.1kJ`

(5)
`H_2O(g)\rightarrow H_2O(l)`
`\Delta H_5=-44.0kJ`

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1)
`2H_2O(g)\rightarrow 2H_2(g)+O_2(g)`
`\Delta H_1=2* 241.8kJ=483.6kJ`

(2)
`XCl_4(s)\rightarrow X(s)+2Cl_2(g)`
`\Delta H_2=-461.9kJ`

(3)
`2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)`
`\Delta H_3=4* -92.3kJ=-369.2kJ`

(4)
`X(s)+O_2(g)\rightarrow XO_2(s)`
`\Delta H_4=-789.1kJ`

(5)
`2H_2O(l)\rightarrow 2H_2O(g)`
`\Delta H_5=2* 44.0kJ=88.0kJ`

The expression for enthalpy of main reaction will be:

`\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5`

`\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)`

`\Delta H=-1048.6kJ`

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ