Question

Calulate \(\Delta S_{(total)}\) for following reaction.
\(\text{CH}_{3}\text{OH}_{(l)}+3/2\text{O}_{2(g)}\rightarrow \text{CO}_{2(g)}+2\text{H}_{2}\text{O}_{(l)}\), \(\Delta H=-726\) kJ mol\({}^{-1}\), \(\Delta S_{(system)}=42\) JK\({}^{-1}\)
(both at \(298\) K)

• A. \(2.478\) kJ K\({}^{-1}\)
• B. \(-2.378\) kJ K\({}^{-1}\)
• C. \(39.58\) kJ K\({}^{-1}\)
• D. \(44.42\) kJ K\({}^{-1}\)

Hint:

Find the entropy change of the surroundings from minus delta H over T, then add the system entropy change.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The total entropy change is \(\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}\).

Step 2: Surroundings entropy change
\(\Delta S_{surr}=-\dfrac{\Delta H}{T}=\dfrac{726000\text{ J mol}^{-1}}{298\text{ K}}=2436.2\text{ J K}^{-1}\).

Step 3: Add the system entropy
\(\Delta S_{total}=42+2436.2=2478.2\text{ J K}^{-1}=2.478\text{ kJ K}^{-1}\).

Step 4: Compare the options
This matches option (A).

Final Answer:
The total entropy change is 2.478 kJ per kelvin. \[ \boxed{\text{(A) }2.478\text{ kJ K}^{-1}} \]