calculate the standard enthalpy of reaction for the following methane–generating reaction of methanogenic bacteria. 4CH3NH2(g) + 2H2O(l)—>3CH4(g) + CO2(g) + 4NH3(g)

To calculate the standard enthalpy of reaction (\(\Delta H^\circ_\text{reaction}\)), we'll use the standard enthalpies of formation (\(\Delta H^\circ_f\)) for the substances involved in the reaction. The general formula is:

\[
\Delta H^\circ_\text{reaction} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants})
\]

### Given Reaction:
\[ 
4 \text{CH}_3\text{NH}_2(g) + 2 \text{H}_2\text{O}(l) \rightarrow 3 \text{CH}_4(g) + \text{CO}_2(g) + 4 \text{NH}_3(g)
\]

### Steps:

1. **List the substances involved and their stoichiometric coefficients:**

   - **Reactants:**
     - \( \text{CH}_3\text{NH}_2(g) \) (4 moles)
     - \( \text{H}_2\text{O}(l) \) (2 moles)
   - **Products:**
     - \( \text{CH}_4(g) \) (3 moles)
     - \( \text{CO}_2(g) \) (1 mole)
     - \( \text{NH}_3(g) \) (4 moles)

2. **Find the standard enthalpy of formation (\(\Delta H^\circ_f\)) for each compound:**

   \[
   \Delta H^\circ_f \text{ (CH}_3\text{NH}_2(g)\text{)} = -22.97 \text{ kJ/mol}
   \]
   \[
   \Delta H^\circ_f \text{ (H}_2\text{O}(l)\text{)} = -285.83 \text{ kJ/mol}
   \]
   \[
   \Delta H^\circ_f \text{ (CH}_4(g)\text{)} = -74.81 \text{ kJ/mol}
   \]
   \[
   \Delta H^\circ_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol}
   \]
   \[
   \Delta H^\circ_f \text{ (NH}_3(g)\text{)} = -45.9 \text{ kJ/mol}
   \]

3. **Apply the formula:**

   \[
   \Delta H^\circ_\text{reaction} = \left[3(-74.81) + 1(-393.5) + 4(-45.9)\right] - \left[4(-22.97) + 2(-285.83)\right]
   \]

4. **Calculate the enthalpy of products:**

   \[
   3(-74.81) + 1(-393.5) + 4(-45.9) = -224.43 - 393.5 - 183.6 = -801.53 \text{ kJ/mol}
   \]

5. **Calculate the enthalpy of reactants:**

   \[
   4(-22.97) + 2(-285.83) = -91.88 - 571.66 = -663.54 \text{ kJ/mol}
   \]

6. **Calculate the standard enthalpy of reaction:**

   \[
   \Delta H^\circ_\text{reaction} = -801.53 \text{ kJ/mol} - (-663.54 \text{ kJ/mol}) = -137.99 \text{ kJ/mol}
   \]

### Final Answer:
The standard enthalpy of reaction is \(-137.99\) kJ/mol.

Would you like more details or have any questions?

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Here are five related questions:

1. How is the standard enthalpy of formation determined experimentally?
2. What role does standard enthalpy play in predicting reaction spontaneity?
3. Can you calculate the entropy change for this reaction?
4. How does temperature affect the standard enthalpy of a reaction?
5. What is the significance of the sign of the enthalpy of reaction?

**Tip:** When dealing with chemical thermodynamics, remember that enthalpy is a state function, which means it depends only on the initial and final states, not the path taken to get there.

Learn how to calculate the standard enthalpy of reaction for the methane-generating reaction of methanogenic bacteria. This problem involves using the standard enthalpies of formation for each substance involved in the reaction, including CH3NH2, H2O, CH4, CO2, and NH3. Suitable for advanced high school or college-level students.

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