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loloroyroy264
14.12.2019 •
Chemistry
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 )
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Ответ:
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:
The intermediate balanced chemical reactions are:
(1)
![\Delta H_1=-241.8kJ](/tpl/images/0418/3281/661e7.png)
(2)
![\Delta H_2=+461.9kJ](/tpl/images/0418/3281/b6ba0.png)
(3)
![\Delta H_3=-92.3kJ](/tpl/images/0418/3281/fb689.png)
(4)
![\Delta H_4=-789.1kJ](/tpl/images/0418/3281/9c593.png)
(5)
![\Delta H_5=-44.0kJ](/tpl/images/0418/3281/235a9.png)
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)
![\Delta H_1=2\times 241.8kJ=483.6kJ](/tpl/images/0418/3281/68fc6.png)
(2)
![\Delta H_2=-461.9kJ](/tpl/images/0418/3281/24fb0.png)
(3)
![\Delta H_3=4\times -92.3kJ=-369.2kJ](/tpl/images/0418/3281/0b2e5.png)
(4)
![\Delta H_4=-789.1kJ](/tpl/images/0418/3281/9c593.png)
(5)
![\Delta H_5=2\times 44.0kJ=88.0kJ](/tpl/images/0418/3281/fd51c.png)
The expression for enthalpy of main reaction will be:
Therefore, the enthalpy of the given reaction will be, -1048.6 kJ
Ответ:
A
Explanation: