Question

calculate the change in entropy for the reaction. 2no₂(g) ⇌ n₂o₄(g)
substance | s° (j/mol·k)
n₂o₄(g) | 304.30
no₂(g) | 240.45
δs° = ? j/k
enter either a + or - sign and the magnitude in your answer.

Explanation:

Step1: Recall the formula for entropy change

The formula for the standard entropy change (\(\Delta S^\circ\)) of a reaction is \(\Delta S^\circ=\sum nS^\circ(\text{products})-\sum mS^\circ(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients.

Step2: Identify products and reactants with coefficients

For the reaction \(2\text{NO}_2(\text{g})
ightleftharpoons \text{N}_2\text{O}_4(\text{g})\), the product is \(\text{N}_2\text{O}_4\) with \(n = 1\), and the reactant is \(\text{NO}_2\) with \(m=2\).

Step3: Substitute values into the formula

\(\Delta S^\circ=S^\circ(\text{N}_2\text{O}_4)- 2\times S^\circ(\text{NO}_2)\)
Substitute \(S^\circ(\text{N}_2\text{O}_4) = 304.30\space\text{J/mol·K}\) and \(S^\circ(\text{NO}_2)=240.45\space\text{J/mol·K}\) into the formula:
\(\Delta S^\circ=304.30-2\times240.45\)

Step4: Calculate the value

First, calculate \(2\times240.45 = 480.9\)
Then, \(\Delta S^\circ=304.30 - 480.9=- 176.6\)

Answer:

\(-176.6\)