Math Problem Statement

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)

Solution

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

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

Given Reaction:

4CH3NH2(g)+2H2O(l)3CH4(g)+CO2(g)+4NH3(g)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:
      • CH3NH2(g)\text{CH}_3\text{NH}_2(g) (4 moles)
      • H2O(l)\text{H}_2\text{O}(l) (2 moles)
    • Products:
      • CH4(g)\text{CH}_4(g) (3 moles)
      • CO2(g)\text{CO}_2(g) (1 mole)
      • NH3(g)\text{NH}_3(g) (4 moles)
  2. Find the standard enthalpy of formation (ΔHf\Delta H^\circ_f) for each compound:

    ΔHf (CH3NH2(g))=22.97 kJ/mol\Delta H^\circ_f \text{ (CH}_3\text{NH}_2(g)\text{)} = -22.97 \text{ kJ/mol} ΔHf (H2O(l))=285.83 kJ/mol\Delta H^\circ_f \text{ (H}_2\text{O}(l)\text{)} = -285.83 \text{ kJ/mol} ΔHf (CH4(g))=74.81 kJ/mol\Delta H^\circ_f \text{ (CH}_4(g)\text{)} = -74.81 \text{ kJ/mol} ΔHf (CO2(g))=393.5 kJ/mol\Delta H^\circ_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol} ΔHf (NH3(g))=45.9 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)=224.43393.5183.6=801.53 kJ/mol3(-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)=91.88571.66=663.54 kJ/mol4(-22.97) + 2(-285.83) = -91.88 - 571.66 = -663.54 \text{ kJ/mol}

  6. Calculate the standard enthalpy of reaction:

    ΔHreaction=801.53 kJ/mol(663.54 kJ/mol)=137.99 kJ/mol\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 137.99-137.99 kJ/mol.

Would you like more details or have any questions?


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.

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Math Problem Analysis

Mathematical Concepts

Chemical Thermodynamics
Enthalpy of Reaction
Enthalpy of Formation

Formulas

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

Theorems

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Suitable Grade Level

Advanced High School or College Level