1. Tardigrade
  2. Question
  3. Chemistry
  4. The heat of combustion of methane at 298 K is expressed by: CH4(g)+2O2(g)→ CO2(g)+2H2O(l) Δ H=890.2 textkJ text. The magnitude of AE of the reaction at this temperature is:

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Q. The heat of combustion of methane at 298 K is expressed by: $ C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l) $ $ \Delta H=890.2\,\text{kJ}\text{.} $ The magnitude of AE of the reaction at this temperature is:

EAMCETEAMCET 1996

Solution:

$ C{{H}_{4}}(g)+2{{O}_{2}}(g)\xrightarrow{{}} $ $ C{{O}_{2}}(g)+2{{H}_{2}}O(l)-890.2\,kJ $ $ \Delta H=\Delta E+\Delta nRT $ $ \Delta n= $ moles of gaseous products - moles of gaseous reactants $ =1-3=-2 $ So, for the above reaction $ \Delta \Epsilon $ value is greater than $ \Delta H. $