Question

Calculate the enthalpy change for the following reaction:$${ CH }_{ 4 }(g)+2{ O }_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g)+2{ H }_{ 2 }O(l)$$Given, enthalpies of formation of $${CH}_{4},{CO}_{2}$$ and $${H}_{2}O$$ are $$74.8\ kJ { mol }^{ -1 },-393.5 \ kJ { mol }^{ -1 },-286\ kJ { mol }^{ -1 }$$ respectively.

Solution

$$\Delta { H }^{ o }=\Delta { H }_{ f(products) }^{ o }-\Delta { H }_{ f(reactants) }^{ o }$$$$=\left[ \Delta { H }_{ ({ CO }_{ 2 }) }^{ o }+\Delta { H }_{ ({ H }_{ 2 }O) }^{ o } \right] -\left[ \Delta { H }_{ ({ CH }_{ 4 }) }^{ o }+2\Delta { H }_{ ({ O }_{ 2 }) }^{ o } \right]$$$$=\left[ -393.5+2\times (-286.2) \right] -\left[ -74.8+2\times 0 \right] =-890.7kJ\quad$$Chemistry

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