Question

Answer please
The above question! ​

Answer 1

Your answer:

$\frac{1}{2}$ N2 + $\frac{3}{2}$ H2 => NH₃

∆S° = product-reactant =

192.6 - ( $\frac{1}{2}$  ×191.5 + $\frac{3}{2}$ × 130.5 ) = 192.6 - 291.5

= -98.9 JK⁻¹ mol⁻¹

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