Question

consider the following intermediate reactions. the overall chemical reaction is as follows. what is the correct enthalpy diagram using the hess law for this system

Answer 1

The question is incomplete, here is the complete question:

Consider the following intermediate reactions.

$CH_4(g)+2O_2\rightarrow CO_2(g)+2H_2O(g);\Delta H_1=-802kJ$

The overall chemical reaction is as follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(I);\Delta H_{rxn}=-890kJ$

What is the correct enthalpy diagram using the Hess law for this system? (Image is attached below)

Answer: The correct image is Image A.

Explanation:

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

According to this law, the enthalpies of these chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to know the required equation.

It also means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given intermediate reactions are:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H_1=-802kJ$                 ...(1)

$2H_2O(g)\rightarrow 2H_2O(l);\Delta H_2=-88kJ$           ....(2)

The final reaction follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H_{rxn}=-890kJ$

Negative sign of the enthalpy represents the heat is being released which is represented by the downward arrow in the diagram.

By adding equation 1 and equation 2, we get:

$\Delta H_{rxn}=\Delta H_1+\Delta H_2\\\\\Delta H_{rxn}=-802+(-88)=-890kJ$

Hence, the correct image is image A.

Answer 2

Answer:

A

Explanation:

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