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What is the heat released or absorbed when 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to form 2 moles of water vapor at constant volume? The enthalpy change for this reaction is -483.7 kJ/mol.

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To calculate the heat released or absorbed in this reaction, we can use the following equation:q = n * Hwhere q is the heat released or absorbed, n is the number of moles of the substance, and H is the enthalpy change per mole.In this case, the balanced chemical equation for the reaction is:2 H g  + O g   2 HO g The enthalpy change  H  for this reaction is given as -483.7 kJ/mol. Since 2 moles of hydrogen gas  H  react with 1 mole of oxygen gas  O  to form 2 moles of water vapor  HO , we can calculate the heat released or absorbed  q  as follows:q = n * Hq = 2 moles *  -483.7 kJ/mol q = -967.4 kJThe negative sign indicates that the reaction is exothermic, meaning that heat is released during the reaction. Therefore, the heat released when 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to form 2 moles of water vapor at constant volume is 967.4 kJ.

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