Question

Calculate the standard entropy change for the reaction p4(g)+5o2(g)→p4o10(s)

Answer 1

give some numerical details .

Answer 2

Answer: The standard entropy change for the reaction is -1067.1 J/K mol.

Explanation:

$P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$

$S^o_{P_4}=41.1 J/K mol$

$S^o_{O_2}=205.2 J/K mol$

$S^o_{P_4O_{10}}=0 J/K mol$(solid)

Standard entropy of the reaction:$\sum(S^o_{products})-\sum(S^o_{reactants})$

Standard entropy of the reaction:

$(S^o_{P_4O_{10}}\times 1+)-(1\times S^o_{P_4}+5\times S^o_{O_2})$

$(0 J/K mol)-(1\times 41.1 J/K mol+5\times 205.2 J/K mol)=-1,067.1 J/K mol$

The standard entropy change for the reaction is -1067.1 J/K mol. Negative sign indicates that energy is released.