What is the change in entropy when 0.290 g of silicon is burned in excess oxygen to yield silicon dioxide at 298 K?
Si(s) + O2(g) -> SiO2(s); ÄS° = –182.4 J/K at 298 K
To find the change in entropy (ΔS) when 0.290 g of silicon is burned in excess oxygen to yield silicon dioxide at 298 K, we need to use the provided standard entropy change (ΔS°) value.
The standard entropy change (ΔS°) is given as -182.4 J/K at 298 K. This value represents the change in entropy when the reaction occurs under standard conditions (1 bar pressure and 298 K temperature) and using the stoichiometric amounts of reactants.
To calculate the actual change in entropy (ΔS) for the given mass of silicon (0.290 g), we can use the equation:
ΔS = ΔS° * (n / n°)
Where:
- ΔS is the actual change in entropy
- ΔS° is the standard entropy change
- n is the number of moles of silicon
- n° is the number of moles of silicon under standard conditions
To find the number of moles of silicon (n), we need to use the molar mass of silicon (28.09 g/mol):
n = mass / molar mass
n = 0.290 g / 28.09 g/mol
Calculating this will give us the number of moles of silicon.
Once we have the number of moles of silicon, we can use it to calculate the actual change in entropy (ΔS) as:
ΔS = -182.4 J/K * (n / n°)
Now, you can substitute the values into the equation and solve for ΔS.