Question

Δs is positive for the reaction ________.

a.2no (g) + o2 (g) → 2no2 (g)
b.2n2 (g) + 3h2 (g) → 2nh3 (g)
c.c3h8 (g) + 5 o2 (g) → 3co2 (g) + 4 h2o (g)
d.mg (s) + cl2 (g) → mgcl2 (s)
e.c2h4 (g) + h2 (g) → c2h6 (g)

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.

Step-by-step 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)`.