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  • The combustion of methane <img alt="\mathrm{\left(\mathrm{CH}_{4}\right) inoxygen \left(\mathrm{O}_{2}\right) producescarbondioxide \left(\mathrm{CO}_{2}\right) andwatervapor \left(\mathrm{H}_{2} \

The combustion of methane \mathrm{\left(\mathrm{CH}_{4}\right) inoxygen \left(\mathrm{O}_{2}\right) producescarbondioxide \left(\mathrm{CO}_{2}\right) andwatervapor \left(\mathrm{H}_{2} \mathrm{O}\right) \cdot Theba}

\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta H=-891.0 \mathrm{~kJ} / \mathrm{mol}
Calculate the standard enthalpy change \left(\Delta H^{\circ}\right) for the combustion of methane.

Option: 1

16.20 \mathrm{~kJ} / \mathrm{mol}


Option: 2

1.270 \mathrm{~kJ} / \mathrm{mol}


Option: 3

-891.0 \mathrm{~kJ} / \mathrm{mol}


Option: 4

20.20 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

The standard enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} for a chemical reaction is the enthalpy change that occurs when one mole of a compound reacts to form its products, all at standard conditions ( 298 \mathrm{~K} and 1 atm pressure). Given the balanced chemical equation:

\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
The given enthalpy change \mathrm{(\Delta H)} is -891.0 \mathrm{~kJ} / \mathrm{mol}. This value is already given at standard conditions, so it directly corresponds to the standard enthalpy change \left(\Delta H^{\circ}\right) for the combustion of methane.

Therefore, the standard enthalpy change \left(\Delta H^{\circ}\right) for the combustion of methane is -891.0 \mathrm{~kJ} / \mathrm{mol}.

Therefore, the correct option is 4 .

Posted by

seema garhwal

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