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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2O(l) → 3CH4(g) + CO2(g) + 4NH3(g) Given that ΔHfo(CH­3NH2, g) = –22.97 kJ/mol; ΔHfo(H2O, l) = –285.8 kJ/mol; ΔHfo(CH4, g) = –74.8 kJ/mol; ΔHfo(CO2, g) = –393.5 kJ/mol ΔHfo(NH3, g) = –46.1 kJ/mol

Respuesta :

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]

For the given chemical reaction:

[tex]4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})][/tex]

We are given:

[tex]\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ[/tex]

Hence, the standard heat for the given reaction is -138.82 kJ

pstnonsonjoku

The heat of reaction is -587.62 kJ/mol.

Given that we have the following information;

  • ΔHfo(CH­3NH2, g) = –22.97 kJ/mol
  • ΔHfo(H2O, l) = –285.8 kJ/mol
  • ΔHfo(CH4, g) = –74.8 kJ/mol
  • ΔHfo(CO2, g) = –393.5 kJ/mol
  • ΔHfo(NH3, g) = –46.1 kJ/mol

Using the formula;

ΔHrxn = ΣΔHproducts - ΣΔHreactants

ΔHrxn = [3(–74.8 ) + 1(–393.5) + 4(–46.1)] - [4(–22.97) + 2(–285.8)]

ΔHrxn =[(-673.2) + (–393.5) + (-184.4)] - [(-91.88) + (-571.6)]

ΔHrxn =(-1251.1) + 663.48

ΔHrxn = -587.62 kJ/mol

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