Consider the exothermic reaction CH4(g) + 2O2(g) -----> CO2(g) + H2O2(g) Calcula

Question

Consider the exothermic reaction CH4(g) + 2O2(g) -----> CO2(g) + H2O2(g) Calculate the standard heat of reaction, or , for this reaction using the given data. Also consider that the standard enthalpy of the formation of elements in their pure form is considered to be zero. Reactant or product CH4(g) -201 kJ/mol CO2(g) -393.5 kJ/mol H2O (g) -241.8 kJ/mol Express your answer to four significant figures and include the appropriate units. im lost anyone able to help its from mastering chemistry chapter 5 i keep trying 0.367 kJ and it says wrong it keeps giving me this hint (Multiply the standard enthalpy change of the reactants and the products by their corresponding stoichiometric coefficients.)

Explanation / Answer

firstly the reaction is a bit incorrectly written.

CH4(g) + 2 O2(g) -----------> CO2(g) + 2 H2O(g)

heat of reation = (Enthalpy of products) - (Enthalpy of reactants)

Enthalpy of products : -393.5 + 2(-241.8) = -393.5 - 483.6 = -877.1 kJ

Enthalpy of reactants : -201 + 2 (0) = -201 kJ (As the enthalpy of oxygen = 0 as it is in it's pure form )

So, finally heat of reaction = -877.1- (-201) = -877.1 + 201 = -676.1kJ

The reaction is a combustion reaction, so heat of reaction must be negative as obtained.

Hope this helped. Please rate it ASAP. Thanks.

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