H2(g)+Cl2(g) →2 HCl(g) ∆H◦ = −185 kJ
Calculate value of ∆Hrxn
a)1mol of HCl is formed
b)3.41g of Cl2 gas reacts
C)result be same if 3.41g H2 reacted
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∆H◦ of a particular reaction corresponds to transformation of 1 mole of reactant/s. In the given example 1 mole H2 (=2 g) reacts with 1 mole Cl2 (=71 g) to produce 2 moles of HCl and 185 kJ heat. Since enthalpy of reaction is an extensive property, i.e. dependant upon the amount, it will be modified depending upon the amount of reactants and products.
Hence in case a), for the production of 1 mole of HCl, enthalpy of reaction shall be 185/2 = -92.5 kJ.
In case b) ∆Hrxn= 185 x 3.41/71 = -8.89 kJ
and in case c) ∆Hrxn= 185 x 3.41/2 = -315.4 kJ.
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