11a. Sodium reacts violently with water according to the equation: 2 Na (s) + 2
ID: 977501 • Letter: 1
Question
11a. Sodium reacts violently with water according to the equation: 2 Na (s) + 2 H2O (l) 2 NaOH (aq) + H2 (g) The resulting solution has a higher temperature than the water prior to the addition of sodium. What are the signs of H° and S° for this reaction?
a) H° is negative and S° is negative.
b) H° is negative and S° is positive.
c) H° is positive and S° is negative.
d) H° is positive and S° is positive.
11 b. Which statement is true for the reaction in part “a” ?
a) The reaction is non-spontaneous at any temperature
b) The reaction is spontaneous at any temperature
c) The reaction is non-spontaneous at lower T, but becomes spontaneous at higher T
d) The reaction is spontaneous at lower T, but becomes non-spontaneous at higher T
11c. Explain your choices for the answers to 11a. and 11b.
Explanation / Answer
11.a
b) Hº is negative and Sº is positive
(I think is easier to consider the changes in enthalpy and entropy due to chemical reaction)
b) deltaH is negative and deltaS is positive for the chemical reaction
(are the same)
11.b
b) the reaction is spontaneous at any temperature
11.c Explanation
The question 11.a says:
"the resulting solution has a higher temperature than the water prior to addition of sodium"
This indicates that the chemical reaction is exothermic, so Hº must be negative.
To determine the sign of Sº, we can do a qualitative analysis of the entropy of the reactants and products of the chemical reaction.
*consider entropy as a unit of disorder, higher values indicates more disorder.
The reactants are sodium in solid phase and water in liquid phase, and the products are sodium hydroxide in solution and hydrogen gas.
As a general rule, consider the increasing order of entropy:
solids < liquids < gases
Since the products are a "liquid" and a gas, and the reactants are a solid and a liquid, is reasonable to assume that the entropy for this chemical reaction, increases. So, Sº is positive.
11.b
To determine the spontaneity of a chemical reaction, we have to consider the next equation:
Gibbs free energy:
deltaG = deltaH - T.deltaS
deltaH, enthalpy change
T, system temperature
delta S, entropy change
For negative values of deltaG, the chemical reaction is considered spontaneous, and since we state in question 11.a that deltaH is negative and deltaS is positive, any input on the equation results in a negative Gibbs free energy. So, the chemical reaction is always spontaneous (although much slower at lower temperatures).
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