Question

The standard entropies of n2, h2 and nh3 are 191.5 , 130.5, 192.6 j/k/mol. the value of delta s of formation of ammonia is

Answer 1

Answer:

-197.8 J/K mol.

Explanation:

The ΔS of formation or the standard entropy of formation is the entropy of the product  - the entropy of the reactant. So, form the question we get that the reaction is  N2 + 3H2 -> 2NH3.

So, we get that 2 moles of NH3 is formed form 1 mole of N2 and 3 moles of H2. So, the entropy of formation will be [2*192.6 - (3*130.5 + 191.5)] which on solving we will get that the entropy of formation will be -197.8 J/K mol.

Answer 2

Answer:-98.9 JK^-1 mol^-1

Explanation:

1/2 N2 + 3/2 H2 => NH3 ∆S° = product-reactant = 192.6 - ( 1/2 ×191.5 + 3/2 × 130.5 ) = 192.6 - 291.5 = -98.9 JK^-1 mol^-1