Question

calculate the enthalpy change for the reaction N2 + 3H2= 2NH3​

Answer 1

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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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