Which of the following reactions has the largest value of ∆S◦ ?
1. K(s) + O2(g) → KO2(s)
2. BaCl2· 2 H2O(s) → BaCl2(s) + 2 H2O(g)
3. N2(g) + 3 H2(g) → 2 NH3(g)
4. NH3(g) + HCl(g) → NH4Cl(s)
5. 2 H2(l) + O2(l) → 2 H2O(g)
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Which of the following reactions has the largest value of ∆S◦ ?
1. K(s) + O2(g) → KO2(s)
2. BaCl2· 2 H2O(s) → BaCl2(s) + 2 H2O(g)
3. N2(g) + 3 H2(g) → 2 NH3(g)
4. NH3(g) + HCl(g) → NH4Cl(s)
5. 2 H2(l) + O2(l) → 2 H2O(g)
Which of the following reactions has the largest value of ∆S◦ ?
1. K(s) + O2(g) → KO2(s)
2. BaCl2· 2 H2O(s) → BaCl2(s) + 2 H2O(g)
3. N2(g) + 3 H2(g) → 2 NH3(g)
4. NH3(g) + HCl(g) → NH4Cl(s)
5. 2 H2(l) + O2(l) → 2 H2O(g)
Correct option (2) BaCl2·2 H2O(s) → BaCl2(s) + 2 H2O(g)
Explanation:
We can predict the sign and magnitude of ∆S by noting the relative order of entropy: solids (lowest) < liquids < solutions < gases (highest) and the number of moles of each type. For the reactions given we have NH3(g) + HCl(g) → NH4Cl(s 2 mol gas → 1 mol solid;
∆S < 0 2 H2(l) + O2(l) → 2 H2O(g)
3 mol liquid → 2 mol gas; ∆S > 0
N2(g) + 3 H2(g) → 2 NH3(g) 4 mol gas → 2 mol gas; ∆S < 0
K(s) + O2(g) → KO2(s) 1 mol solid + 1 mol gas → 1 mol solid; ∆S < 0
BaCl2 · 2 H2O(s) → BaCl2(s) + 2 H2O(g)
1 mol solid → 1 mol solid + 1 mol gas; ∆S > 0
The greatest increase in S would be for the reaction
BaCl2· 2 H2O(s) → BaCl2(s) + 2 H2O(g)