# Question 2d5bf

Sep 1, 2017

You use the standard molar entropies ${S}^{\circ}$ of the reactants and products.

#### Explanation:

The formula to use is

color(blue)(barul|stackrel(" ")(Δ_text(rxn)S^@ = sumnS_text(prod)^@ - summS_text(react)^@)|)

where $m$ and $n$ are the moles of the reactants and products in the balanced equation.

Here's how to work the first example.

$\textcolor{w h i t e}{m m m m m m m m m} \text{C(s)" + "H"_2"O(g)" → "CO(g)" + "H"_2"(g)}$
${S}^{\circ} \text{/J·K"^"-1""mol"^"-1} : \textcolor{w h i t e}{m} 5.694 \textcolor{w h i t e}{m l} 188.72 \textcolor{w h i t e}{m m l} 197.91 \textcolor{w h i t e}{m l} 130.59$

Δ_text(rxn)S^@ = "(1×197.91 + 1 × 130.59) - (1×5.694 + 1 ×188.72) J/K" = "134.09 J/K"#

The other two reactions use the same method.