Question

If the enthalpy change for the following reaction at $300\,\text{K}$ is $+7\,\text{kJ mol}^{-1}$, find the entropy change of the surroundings.
$H_2O_{(s)} \longrightarrow H_2O_{(l)}$

• A. $-42.8~J~K^{-1}$
• B. $-23.3~J~K^{-1}$
• C. $-30.7~J~K^{-1}$
• D. $-110.0~J~K^{-1}$

Hint:

Thermodynamics Tip: Always remember the negative sign! An endothermic process ($\Delta H>0$) absorbs heat from the surroundings, thereby decreasing the entropy of the surroundings ($\Delta S_{surr}<0$).

Answer 1

The Correct Option is B

Concept: Chemistry (Thermodynamics) - Entropy Change of Surroundings.

Step 1: Identify the given thermodynamic parameters. The absolute temperature of the process is given as $T = 300~K$. The enthalpy change of the system (melting of ice) is given as $\Delta H_{sys} = +7~kJ~mol^{-1}$.

Step 2: Convert units to standard Joules. Since the given multiple-choice options are in Joules per Kelvin ($J~K^{-1}$), we must convert the system's enthalpy change from kilojoules to joules: $\Delta H_{sys} = +7000~J~mol^{-1}$.

Step 3: State the formula for entropy change of the surroundings. At constant temperature and pressure, the heat transferred to the surroundings is equal and opposite to the heat absorbed by the system. Thus, $q_{surr} = -\Delta H_{sys}$. The formula for the entropy change of the surroundings is $\Delta S_{surr} = \frac{q_{surr}}{T} = \frac{-\Delta H_{sys}}{T}$.

Step 4: Substitute the values into the formula. Substitute the converted enthalpy value and the given temperature into the equation: $\Delta S_{surr} = \frac{-7000~J~mol^{-1}}{300~K}$.

Step 5: Calculate the final numerical value. Cancel the zeros and perform the division: $\Delta S_{surr} = \frac{-70}{3} = -23.333...~J~K^{-1}mol^{-1}$. Rounding to one decimal place to match the provided options yields $-23.3~J~K^{-1}mol^{-1}$. $$ \therefore \text{The entropy change of the surroundings is approximately } -23.3~J~K^{-1}. $$