Question

Consider the following intermediate chemical equations. 3 equations. 1: upper C upper H subscript 4 (g) right arrow upper C (s) plus 2 upper H subscript 2 (g). Delta H 1 equals 74. 8 kilojoules. 2: upper C upper C l subscript 4 (g) right arrow upper C (s) plus 2 upper C l subscript 2 (g). Delta H 2 equals 95. 7 kilojoules. Upper H subscript 2 (g) plus upper C l subscript 2 (g) right arrow 2 upper H upper C l (g). Delta H subscript 3 equals negative 92. 3 kilojoules. What is the enthalpy of the overall chemical reaction Upper C upper H subscript 4 (g) plus 4 upper C l subscript 2 (g) right arrow upper C upper C l subscript 4 (g) plus 4 upper H upper C l (g). ? -205. 7 kJ -113. 4 kJ -14. 3 kJ 78. 0 kJ.

Answer 1

The enthalpy of the reaction is the amount of energy absorbed or lost in the reaction. The enthalpy of the final reaction  is -113.4 kJ.

What are intermediate reactions?

The intermediate reactions are the small reactions that takes place in order to achieve the final reaction product.

The given intermediate reactions are:

$\rm CH_4\;\rightarrow\;C\;+\;2\;H_2......\Delta H_1=74. 8\;kJ\\CCl_4\;\rightarrow\;C\;+\;2\;Cl_2......\Delta H_2=95. 7\;kJ\\H_2\;+\;Cl_2\;\rightarrow\;2\;HCl......\Delta H_3=-92. 3\;kJ$

The final reaction to be achieved is:

$\rm CH_4\;+\;4\;Cl_2\;\rightarrow\;CCl_4\;+\;4\;HCl$

The second reaction is reversed in order to achieve the final reaction product.

The enthalpy of the final reaction ($\Delta H_{rxn}$) is given as:

$\Delta H_{rxn}=\Delta H_1-\Delta H_2+\Delta H_3$

Substituting the values of enthalpy:

$\Delta H_{rxn}= 74. 8-95. 7+(-92. 3)\text k\text J\\\Delta H_{rxn}=-113.4\;\rm kJ$

The final enthalpy of the reaction is -113.4 kJ. Thus, option B is correct.

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