Respuesta :
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: 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 [tex]\Delta S[/tex]. More is the degree of randomness present more positive will be the value of [tex]\Delta S[/tex]. Similarly, less is the degree of randomness present within a substance lesser will be the value of [tex]\Delta S[/tex].
(a) [tex]2NO(g) + O_{2}(g) \rightarrow 2NO_{2}(g)[/tex]
Here, 3 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.
(b) [tex]2N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)[/tex]
Here, 5 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.
(c) [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex]
Here, 6 moles of reactants are giving 7 moles of product. Hence, entropy is increasing so, the value of [tex]\Delta S[/tex] is positive.
(d) [tex]Mg(s) + Cl_{2}(g) \rightarrow MgCl_{2}(s)[/tex]
Here, 2 moles of reactants are giving 1 mole of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.
(e) [tex]C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)[/tex]
Here, 2 moles of reactants are giving 1 mole of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.
Thus, we can conclude that [tex]\Delta S[/tex] is positive for the reaction [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex].
