at its melting point, O c, the enthalpy of fusion of water is 1.435 kcal/mol. what is the molar entropy change for the melting ice at 0C?
answer: 5.26 cal/mol-k
dG = dH - TdS
At the melting point of zero C the system is in equilibrium; therefore, dG = 0. Substitute and solve for dS.
so dH/T?
To find the molar entropy change for the melting ice at 0°C, we can use the equation:
∆S = ∆H / T
where:
∆S is the molar entropy change
∆H is the enthalpy of fusion of water at its melting point
T is the temperature in Kelvin
First, we need to convert the enthalpy of fusion from kilocalories per mole (kcal/mol) to calories per mole (cal/mol) by multiplying by 1000 since there are 1000 calories in 1 kilocalorie:
∆H = 1.435 kcal/mol × 1000 cal/kcal = 1435 cal/mol
Next, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 0°C + 273.15 = 273.15 K
Now, we can plug these values into the equation:
∆S = 1435 cal/mol / 273.15 K = 5.26 cal/mol-K
Therefore, the molar entropy change for the melting ice at 0°C is 5.26 cal/mol-K.