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Consider the following intermediate reactions:

[tex]\[
\begin{array}{ll}
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) & \Delta H_1 = -802 \, \text{kJ} \\
2 H_2O(g) \rightarrow 2 H_2O(l) & \Delta H_2 = -88 \, \text{kJ}
\end{array}
\][/tex]

The overall chemical reaction is as follows:

[tex]\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -890 \, \text{kJ}
\][/tex]

What is the correct enthalpy diagram using Hess's law for this system?

[tex]\[
\begin{array}{l|c|r}
CH_4(g) + 2 O_2(g) & & \\
\hline
& & \\
& & -802 \, \text{kJ} \\
& & \\
& & -88 \, \text{kJ} \\
& & \\
\hline
& & -890 \, \text{kJ}
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's approach this problem step by step, utilizing Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the intermediate steps that lead to the overall reaction.

### Given Reactions and Enthalpy Changes:
1. Intermediate Reaction 1:
[tex]\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \quad \Delta H_1 = -802\, \text{kJ} \][/tex]

2. Intermediate Reaction 2:
[tex]\[ 2H_2O(g) \rightarrow 2H_2O(l) \quad \Delta H_2 = -88\, \text{kJ} \][/tex]

3. Overall Reaction:
[tex]\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_{\text{overall}} = -890\, \text{kJ} \][/tex]

### Using Hess's Law:

To determine the correctness of the enthalpy diagram, we sum the enthalpy changes of the intermediate reactions and ensure they match the overall enthalpy change for the reaction.
[tex]\[ \Delta H_{\text{calculated overall}} = \Delta H_1 + \Delta H_2 = -802\, \text{kJ} + (-88\, \text{kJ}) = -890\, \text{kJ} \][/tex]

### Check for Correctness:
We see that:
[tex]\[ \Delta H_{\text{calculated overall}} = -890\, \text{kJ} = \Delta H_{\text{overall}} \][/tex]
This confirms the enthalpy diagram is correct according to Hess's law.

### Enthalpy Diagram:

The enthalpy diagram represents the changes in enthalpy for the reactions in a visual manner. It can be sketched as follows:

1. Start with [tex]\(CH_4(g) + 2O_2(g)\)[/tex]:
- The initial enthalpy level for [tex]\(CH_4(g) + 2O_2(g)\)[/tex].

2. After the first reaction:
- The system releases [tex]\(-802\, \text{kJ}\)[/tex], dropping to the enthalpy level of [tex]\(CO_2(g) + 2H_2O(g)\)[/tex].

3. After the second reaction:
- The system releases an additional [tex]\(-88\, \text{kJ}\)[/tex], reaching the final enthalpy level of the overall reaction [tex]\(CO_2(g) + 2H_2O(l)\)[/tex].

This can be summarized in the following diagram:

[tex]\[ \begin{array}{rcl} CH_4(g) + 2O_2(g) & & \\ & \downarrow & \text{(-802 kJ)} \\ CO_2(g) + 2H_2O(g) & & \\ & \downarrow & \text{(-88 kJ)} \\ CO_2(g) + 2H_2O(l) & & \\ \end{array} \][/tex]

Hence, the enthalpy diagram is:

[tex]\[ \begin{array}{rcl} CH_4(g) + 2O_2(g) & \quad & \\ & & \\ -802 \text{kJ} & & \\ & & -890\, \text{kJ} \\ CO_2(g) + 2H_2O(g) & & \\ & & -88 \text{kJ} \\ CO_2(g) + 2H_2O(l) & & \\ \end{array} \][/tex]

This diagram correctly represents the enthalpy changes for the intermediate and overall reactions, confirming the validity of the given enthalpy values.
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