Question

Calculate \(\Delta \text{S}_{\text{total}}\) for a certain reaction at \(298\text{ K}\) if \(\Delta \text{H}^\circ = -208.6\text{ kJ}\) and \(\Delta \text{S}^\circ = -36\text{ J K}^{-1}\)

• A. \(664\text{ J K}^{-1}\)
• B. \(834\text{ J K}^{-1}\)
• C. \(926\text{ J K}^{-1}\)
• D. \(736\text{ J K}^{-1}\)

Hint:

Always convert \(\Delta H\) into Joules before calculation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The total entropy change of the universe (\(\Delta S_{\text{total}}\)) for a process is the sum of the entropy change of the system (\(\Delta S_{\text{sys}}\)) and the entropy change of its surroundings (\(\Delta S_{\text{surr}}\)).
Step 2: Key Formula or Approach:
\[ \Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \] The entropy change of the surroundings is related to the enthalpy change of the system by the equation: \[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} \] Step 3: Detailed Explanation:
Given values:
Temperature, \(T = 298\text{ K}\)
Enthalpy change of system, \(\Delta H_{\text{sys}} = \Delta H^\circ = -208.6\text{ kJ} = -208600\text{ J}\) (converted to Joules to match units)
Entropy change of system, \(\Delta S_{\text{sys}} = \Delta S^\circ = -36\text{ J K}^{-1}\)
First, calculate \(\Delta S_{\text{surr}}\): \[ \Delta S_{\text{surr}} = -\frac{-208600\text{ J}}{298\text{ K}} \] \[ \Delta S_{\text{surr}} = +\frac{208600}{298}\text{ J K}^{-1} = +700\text{ J K}^{-1} \] Now, calculate \(\Delta S_{\text{total}}\): \[ \Delta S_{\text{total}} = -36\text{ J K}^{-1} + 700\text{ J K}^{-1} \] \[ \Delta S_{\text{total}} = 664\text{ J K}^{-1} \] Step 4: Final Answer:
The total entropy change is \(664\text{ J K}^{-1}\), which corresponds to option (A).