Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Consider the chemical equations shown here:

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

Which equation shows how to calculate [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the equation below?

[tex]\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)
\][/tex]

Sagot :

GinnyAnswer
To calculate the enthalpy change, [tex]\( \Delta H_{\text{rxn}} \)[/tex], for the reaction:

[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O (l), \][/tex]

we need to combine the enthalpy changes of the given reactions. Here are the two reactions provided:

1. [tex]\( CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \)[/tex], with [tex]\( \Delta H_1 = -802 \text{ kJ} \)[/tex]
2. [tex]\( 2 H_2O(g) \rightarrow 2 H_2O(l) \)[/tex], with [tex]\( \Delta H_2 = -88 \text{ kJ} \)[/tex]

The first reaction represents the combustion of methane yielding carbon dioxide and water in the gaseous state. The second reaction represents the phase change of water from gas to liquid.

To find the overall [tex]\( \Delta H_{\text{rxn}} \)[/tex] for the formation of liquid water directly from the combustion of methane, we sum the enthalpy changes of the individual steps:

[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 \][/tex]

Given the values:

[tex]\[ \Delta H_1 = -802 \text{ kJ} \][/tex]
[tex]\[ \Delta H_2 = -88 \text{ kJ} \][/tex]

Add these together:

[tex]\[ \Delta H_{\text{rxn}} = -802 \text{ kJ} + -88 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -890 \text{ kJ} \][/tex]

Thus, the equation to calculate [tex]\( \Delta H_{\text{rxn}} \)[/tex] for the reaction:

[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O (l) \][/tex]

is:

[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 = -802 \text{ kJ} + -88 \text{ kJ} = -890 \text{ kJ} \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.