Question

Δs is positive for the reaction ________.

Answer 1

C. The reaction goes from one of less moles to more causing it to be more disordered and therefore have a positive change in s

Answer 2

Answer: Option (c) is the correct answer.

Explanation:

Entropy means the degree of randomness present in a substance or within the reactants in a chemical reaction.

Change in entropy is represented by $\Delta S$. More is the degree of randomness present more positive will be the value of $\Delta S$. Similarly, less is the degree of randomness present within a substance lesser will be the value of $\Delta S$.

(a)    $2NO(g) + O_{2}(g) \rightarrow 2NO_{2}(g)$

Here, 3 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

(b)   $2N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)$

Here, 5 moles of reactants are giving 2 moles of product.  Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

(c)   $C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)$

Here, 6 moles of reactants are giving 7 moles of product.  Hence, entropy is increasing so, the value of $\Delta S$ is positive.

(d) $Mg(s) + Cl_{2}(g) \rightarrow MgCl_{2}(s)$

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

(e)  $C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)$

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

Thus, we can conclude that $\Delta S$ is positive for the reaction $C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)$.