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At 27^@C latent heat of I^- fusion of a ...

At `27^@C` latent heat of `I^-` fusion of a compound is `2.7xx10^3 J mol^(-1)`. Calculate the entropy change during fusion.

A

`9 JK^(-1) "mol"^(-1)`

B

`10.73 JK^(-1)"mol"^(-1)`

C

`2930 JK^(-1)"mol"^(-1)`

D

`108.5 JK^(-1)"mol"^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the entropy change during the fusion of a compound, we can use the formula: \[ \Delta S = \frac{Q}{T} \] where: - \(\Delta S\) is the entropy change, - \(Q\) is the latent heat of fusion (in joules per mole), - \(T\) is the absolute temperature (in kelvin). ### Step 1: Convert the given temperature to Kelvin The temperature is given as \(27^\circ C\). To convert this to Kelvin, we use the formula: \[ T(K) = T(°C) + 273.15 \] Calculating: \[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \] ### Step 2: Identify the latent heat of fusion The latent heat of fusion given in the problem is: \[ Q = 2.7 \times 10^3 \, J/mol \] ### Step 3: Calculate the entropy change Now, we can substitute the values of \(Q\) and \(T\) into the entropy change formula: \[ \Delta S = \frac{Q}{T} = \frac{2.7 \times 10^3 \, J/mol}{300 \, K} \] Calculating: \[ \Delta S = \frac{2700 \, J/mol}{300 \, K} = 9 \, J/K \cdot mol \] ### Final Answer The entropy change during fusion is: \[ \Delta S = 9 \, J/K \cdot mol \]
To calculate the entropy change during the fusion of a compound, we can use the formula: \[ \Delta S = \frac{Q}{T} \] where: - \(\Delta S\) is the entropy change, ...
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