Question

Calculate ΔS_rxn for the following chemical equation: NaOH(aq) + HCl(aq) → H₂O(l) + NaCl(aq). Round to the nearest whole number. Given entropy values: S_NaOH = 49.8 J/mol K, S_HCl = 56.5 J/mol K, S_H₂O = 69.91 J/mol K, S_NaCl = 115.5 J/mol K.

a) -35 J/mol K
b) -10 J/mol K
c) 35 J/mol K
d) 10 J/mol K

Answer 1

The calculated entropy change, ΔSrxn, for the given reaction using the formula ΔSrxn = (∑Sproducts) - (∑Sreactants) and the provided entropy values is 79 J/mol K, which doesn't match any of the answer choices provided.

Step-by-step explanation:

To calculate the ΔSrxn (entropy change) for the chemical equation NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq), you use the standard entropy values for each substance provided and employ the following formula:

ΔSrxn = (∑Sproducts) - (∑Sreactants)

Given entropy values:

• SNaOH = 49.8 J/mol K
• SHCl = 56.5 J/mol K
• SH2O = 69.91 J/mol K
• SNaCl = 115.5 J/mol K

Apply the values to the formula:

ΔSrxn = (SH2O + SNaCl) - (SNaOH + SHCl)
ΔSrxn = (69.91 J/mol K + 115.5 J/mol K) - (49.8 J/mol K + 56.5 J/mol K)
ΔSrxn = 185.41 J/mol K - 106.3 J/mol K
ΔSrxn = 79.11 J/mol K

Round to the nearest whole number:

ΔSrxn = 79 J/mol K

The correct answer is not included in the options provided, as the calculated value is 79 J/mol K and not any of the given options (a) -35 J/mol K, (b) -10 J/mol K, (c) 35 J/mol K, or (d) 10 J/mol K.