Question

Scoring: Your score will be based on the number of correct matches minus the number of incorrect matches. There is no penalty for missing matches. Use the References to access important values if needed for this question. Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All H2(g) + F2(g)2HF(g) 2SO3(g)2SO2(g) + O2(g) CH4(g) + 2O2(g)CO2(g) + 2H2O(g) 2H2S(g) + 3O2(g)2H2O(g) + 2SO2(g) 2H2O2(l)2H2O(l) + O2(g)

Answer 1

Answer:

a. Too close to zero

b. ΔS > 0

c. Too close to zero

d. ΔS < 0

e.  ΔS > 0

Explanation:

In order to answer this question we need to compare the change in number of mol in the gas state for the products vs the reactants.

An increase in number moles gas leads to a positive change in entropy. Conversely a decrease will mean ΔS is negative.

Now we can solve the parts in this question.

a. H₂ (g) + F₂ (g)   ⇒  2 HF (g)

We have no change in the number of moles gas products minus reactants. Therefore is too close to zero to decide since there is no change in mol gas.

b. 2 SO₃ ( g) ⇒ 2 SO₂ (g) + O₂ (g)

We have three moles of gas products starting with 2 mol gas SO₂, therefore ΔS is positive.

c. CH₄ (g) + 2 O₂ (g)   ⇒ CO₂ (g) + 2 H₂O (g)

Again, we have the same number of  moles gas in the products and the reactants (3), and it is too close to zero to decide

d.  2 H₂S (g) + 3 O₂(g) ⇒ 2 H2O (g) + 2 SO₂ (g)

Here the change in number of moles gas is negative ( -1 ), so will expect ΔS < 0

e. 2H₂O₂ (l)   ⇒ 2H₂O(l) +- O2(g)

Here we have 1 mol gas and 2 mol liquid  produced from 2 mol liquid reactants, thus the change in entropy is positive.