Question:
What is the change in entropy of surrounding for the reaction,
\(\text{H}_{2(g)}+1/2 \text{O}_{2(g)}\rightarrow \text{H}_2\text{O}_{(l)}\)
at 298 K if standard enthalpy of formation of water is \(-286\) kJ ?
What is the change in entropy of surrounding for the reaction,
\(\text{H}_{2(g)}+1/2 \text{O}_{2(g)}\rightarrow \text{H}_2\text{O}_{(l)}\)
at 298 K if standard enthalpy of formation of water is \(-286\) kJ ?
\(\text{H}_{2(g)}+1/2 \text{O}_{2(g)}\rightarrow \text{H}_2\text{O}_{(l)}\)
at 298 K if standard enthalpy of formation of water is \(-286\) kJ ?
Show Hint
Surroundings gain the heat released: delta S_surr = -delta H / T.
Updated On: Oct 1, 2026
- \(959.7 \text{JK}^{-1}\)
- \(801.5 \text{JK}^{-1}\)
- \(850.7 \text{JK}^{-1}\)
- \(980.0 \text{JK}^{-1}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The heat given out by the reaction is absorbed by the surroundings at constant temperature and pressure. So the entropy change of the surroundings is tied to \(\Delta H\) of the reaction.
Step 2: Key Formula or Approach:
\[ \Delta S_{surr} = -\frac{\Delta H}{T} \]
Step 3: Detailed Explanation:
The reaction forms 1 mol of liquid water, so \(\Delta H = \Delta_f H^\circ = -286\) kJ = \(-286000\) J.
\[ \Delta S_{surr} = -\frac{-286000}{298} = +959.7\ \text{J K}^{-1} \]
The value is positive because the surroundings absorb heat from an exothermic reaction. Options (B), (C) and (D) do not result from dividing 286000 J by 298 K.
Step 4: Check:
\(286000/298 = 959.73\), which matches option (A).
Final Answer:
Surroundings gain 286 kJ of heat at 298 K. \[ \boxed{\text{(A) }959.7\ \text{J K}^{-1}} \]
The heat given out by the reaction is absorbed by the surroundings at constant temperature and pressure. So the entropy change of the surroundings is tied to \(\Delta H\) of the reaction.
Step 2: Key Formula or Approach:
\[ \Delta S_{surr} = -\frac{\Delta H}{T} \]
Step 3: Detailed Explanation:
The reaction forms 1 mol of liquid water, so \(\Delta H = \Delta_f H^\circ = -286\) kJ = \(-286000\) J.
\[ \Delta S_{surr} = -\frac{-286000}{298} = +959.7\ \text{J K}^{-1} \]
The value is positive because the surroundings absorb heat from an exothermic reaction. Options (B), (C) and (D) do not result from dividing 286000 J by 298 K.
Step 4: Check:
\(286000/298 = 959.73\), which matches option (A).
Final Answer:
Surroundings gain 286 kJ of heat at 298 K. \[ \boxed{\text{(A) }959.7\ \text{J K}^{-1}} \]
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