calculate the work done when 50.0g of tin dissolves in excess acid at 1.00 atm and 25 degrees c.
To calculate the work done when 50.0g of tin dissolves in excess acid at 1.00 atm and 25 degrees C, we need to determine the change in Gibbs free energy (ΔG) and use the equation:
Work = -ΔG
First, we need to find the change in Gibbs free energy for the dissolution of tin. We can use the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change and ΔS is the entropy change.
The enthalpy change (ΔH) can be determined using the molar enthalpy of dissolution of tin (ΔH_soln), and the number of moles of tin dissolved.
1. Convert the mass of the tin (50.0g) to moles:
molar mass of tin (Sn) = 118.71 g/mol (rounded to 2 decimal places)
moles of tin = mass / molar mass = 50.0g / 118.71 g/mol = 0.42 mol (rounded to 2 decimal places)
2. Use the molar enthalpy of dissolution for tin (ΔH_soln):
ΔH_soln of tin = -9.7 kJ/mol (rounded to 1 decimal place)
ΔH = ΔH_soln * moles of tin
ΔH = -9.7 kJ/mol * 0.42 mol = -4.1 kJ (rounded to 1 decimal place)
Next, we need to calculate the change in entropy (ΔS) for the dissolution of tin.
The change in entropy is related to the number of gas molecules formed or consumed during the reaction. In this case, no gas is produced or consumed, so the change in entropy is likely very small.
ΔS = ΔS_sys + ΔS_surr
For the dissolution of tin, the system does not produce or consume any gas molecules, so ΔS_sys is negligible.
ΔS_surr = -ΔH / T
T = 25 + 273.15 = 298.15 K
ΔS = -4.1 kJ / 298.15 K = -0.014 J/K (rounded to 3 decimal places)
Now that we have the ΔH and ΔS, we can calculate the change in Gibbs free energy (ΔG):
ΔG = ΔH - TΔS
ΔG = -4.1 kJ - (298.15 K * -0.014 J/K) = -4.1 kJ + 4.2 J = -4.1 kJ + 0.0042 kJ = -4.0958 kJ (rounded to 4 decimal places)
Finally, the work done (W) is equal to the negative of ΔG:
Work = -ΔG = -(-4.0958 kJ) = 4.0958 kJ (rounded to 4 decimal places)
Therefore, the work done when 50.0g of tin dissolves in excess acid at 1.00 atm and 25 degrees C is approximately 4.0958 kJ.
To calculate the work done when tin dissolves in excess acid, we need to use the equation:
Work = -PΔV
where:
Work is the work done (in Joules, J)
P is the pressure (in atmospheres, atm)
ΔV is the change in volume (in liters, L)
To compute the work, we need to determine the change in volume.
First, we need to convert the mass of tin into moles. We can use the molar mass of tin (Sn) to do this. The molar mass of tin is 118.71 g/mol.
Mass of tin = 50.0 g
Molar mass of tin = 118.71 g/mol
By dividing the mass of tin by its molar mass, we can determine the number of moles of tin:
Number of moles of tin = mass of tin / molar mass of tin
= 50.0 g / 118.71 g/mol
Next, we need to calculate the volume change using the ideal gas law:
PV = nRT
where:
P is the pressure (in atm)
V is the volume (in L)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (in Kelvin, K)
We know the pressure is 1.00 atm and the temperature is 25 degrees Celsius. To convert to Kelvin, we add 273 to the Celsius temperature:
T in Kelvin = 25 + 273
= 298 K
Now, we can rearrange the ideal gas law equation to solve for the change in volume:
ΔV = nRT / P
Substituting the values we have:
ΔV = (number of moles) * (ideal gas constant) * (temperature) / pressure
= (number of moles of tin) * (0.0821 L·atm/(mol·K)) * (298 K) / (1.00 atm)
Finally, to calculate the work done:
Work = -PΔV
= -(1.00 atm) * (ΔV)
Plug in the value of ΔV and calculate the work done.