Question

Given the standard entropy values in the table, what is the value of `Delta"Sº"_"reaction"` for the following reaction?

Answer 1

Answer:

$\Delta S^o=-149\frac{J}{mol*K}$

Explanation:

Hello,

In this case, we define the standard change in the entropy for a chemical reaction as:

$\Delta S^o=\Sigma \nu _iS_i^o$

Whereas $\nu _i$ is the stoichiometric coefficient of the species i which is negative for products and positive for reactants and $S_i^o$ is the standard entropy for the species i in J/mol*K. Thus, we have:

$\Delta S^o=(-2)(53} )+(-3)(205)+(2)(248)+(2)(38)\\\Delta S^o=-149\frac{J}{mol*K}$

Best regards.

Answer 2

Answer:

entropy of reaction = 572-718= -146J/mol-k.

Explanation:

entropy of a reaction = entropy of product - entropy of reactant.

entropy of product = 2×entropy of SO2+2×entropy of NiO

                               =$2\times 248+2\times 38$

                              =572 J/mol-k

entropy of reactant = 2×entropy if NiS+3×entropy of O2

                                =$2\times 53+3\times 205$

                                =718 J/mol-k

therefore entropy of reaction = 572-718= -146J/mol-k.