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What is the standard enthalpy change for the reaction N2(g) + 3H2(g) → 2NH3(g) at 298 K, given that the standard enthalpy of formation of NH3(g) is -46.1 kJ mol-1?

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To calculate the standard enthalpy change for the reaction, we can use the following equation:H reaction  =  Hf products  -  Hf reactants In this case, the reactants are N2 g  and H2 g , and the product is NH3 g . The standard enthalpy of formation for elements in their standard states  N2 and H2  is zero. Therefore, the equation becomes:H reaction  = [2  Hf NH3 ] - [Hf N2  + 3  Hf H2 ]Given that the standard enthalpy of formation of NH3 g  is -46.1 kJ mol-1:H reaction  = [2   -46.1 kJ mol-1 ] - [0 + 3  0]H reaction  = -92.2 kJ mol-1So, the standard enthalpy change for the reaction N2 g  + 3H2 g   2NH3 g  at 298 K is -92.2 kJ mol-1.

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