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NEETChemistryThermodynamics (C)

Consider the thermochemical equation for the synthesis of ammonia: N ₂( g ) + 3 H ₂( g ) 2 NH ₃( g ) _ r H ^ = -92.4 ~kJ What is the standard enthalpy change for the decomposition of one mole of ammonia gas into its constituent elements?

Options

  1. A-46.2 ~kJ
  2. B+92.4 ~kJ
  3. C-92.4 ~kJ
  4. D+46.2 ~kJ

Correct answer

D. +46.2 ~kJ

Step-by-step solution

The given reaction is the formation of 2 moles of NH ₃ with _ r H ^ = -92.4 ~kJ . The decomposition of ammonia is the reverse reaction: 2 NH ₃( g ) N ₂( g ) + 3 H ₂( g ) For this reverse reaction, the sign of the enthalpy change is reversed: H ^ = +92.4 ~kJ for the decomposition of 2 moles of NH ₃ . For the decomposition of 1 mole of NH ₃ , the enthalpy change is: H ^ = +92.4 2 = +46.2 ~kJ . Answer: +46.2 ~kJ

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