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What is the change in internal energy of a system when two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water vapor at constant temperature and pressure?

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To calculate the change in internal energy  U  of a system, we can use the equation:U = q + wwhere q is the heat absorbed or released by the system, and w is the work done on or by the system.In this case, we have a constant temperature and pressure reaction of hydrogen gas  H2  and oxygen gas  O2  forming water vapor  H2O :2 H2 g  + O2 g   2 H2O g Since the reaction is occurring at constant temperature and pressure, we can use the enthalpy change  H  as a proxy for the heat absorbed or released  q  in the reaction. The enthalpy change for this reaction can be found using the standard enthalpies of formation  Hf  of the products and reactants:H =  Hf products  -  Hf reactants The standard enthalpy of formation for H2 g  and O2 g  is 0 kJ/mol, as they are in their standard states. The standard enthalpy of formation for H2O g  is -241.8 kJ/mol. Using these values, we can calculate the enthalpy change for the reaction:H = [2   -241.8 kJ/mol ] - [2  0 kJ/mol + 1  0 kJ/mol] = -483.6 kJNow, we need to consider the work done on or by the system. Since the reaction is occurring at constant pressure, we can use the equation:w = -PVwhere P is the pressure and V is the change in volume. However, since we are given no information about the pressure or volume changes, we cannot calculate the work done on or by the system.Without the work  w  value, we cannot accurately calculate the change in internal energy  U  of the system. However, we can still conclude that the reaction is exothermic, as the enthalpy change  H  is negative, meaning heat is released during the reaction.

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