Question

The enthalpy change (ΔH) for the reaction, N2(g) + 3H2(g) → 2NH3(g) is - 92.38 kJ at 298 K. What is ΔE at 298 K?

Answer 1

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 Method

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....

Answer 2

For isothermal process;

=delta E = delta H - delta nRT

= -92.38 - ( -2×8.31×298 / 1000)

-92,38 + 4.95

-87.43 KJ