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Calculate the enthalpy change for the reaction N2 + 3H2= 2NH3 |
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Answer» Answer: mark as brainliest and follow me plz Explanation: ✅✅N2 (g) + 3H2 (g) ====》 2NH3 (g) ( total moles of gaseous reactant) > ( total moles of gaseous PRODUCTS) :: ( Disordered state of gaseous reactant) > ( Disordered state of gaseous products) . In the reaction, 4 mole of gaseous reactant from 2 moles of gaseous products, I.e, DELTA n > 0. Therefore disorder DECREASES & HENCE entropy decreases delta S < 0.. ✔For isothermal process; =delta E = delta H - delta nRT = -92.38 - ( -2×8.31×298 / 1000) -92,38 + 4.95 -87.43 KJ... ✔✔ other methods. delta ng = 2.4 = -2 delta H = delta U + delta ngRT delta U = delta H - delta ngR == - 92.38 × 1000 - (-2) × 8.314 × 298 == - 87.424KJ.... .... |
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