Calculate the expansion work done when 50 g of water is electrolysed under constant pressure at 25C
Here is the answer
2H2O -------> O2 + 2H2
2.7 mol 1,38 mol 2,7 mol
mol H2O = 50g/18 g/mol = 2.7 mol
for general eq -> w = - Pex.delta V
Pex= external pressure
delta V = Vfinal (V2) - V initial(V1)
neglect the intial volume because the final volume (after the production of gas) is so much larger and delta V = Vf- Vi=Vf=nRT/Pex, then : W = -(n.R.T)
wO2 = -(nxRxT)
wO2 = 1,38 mol x 8,3145 J.K^-1.mol^-1 x 298 =-3,4 Kj
same for calcualting O2 then wH2 = -6,8 kJ
Wtotal = WO2 + WH2 = -3,4 Kj + (-6,8 Kj) = 10,2 Kj or 10 Kj
-10KJ
-10KJ
-10287 J
how did u get -10 kj
Please show procedures
10287 J
To calculate the expansion work done during the electrolysis process, we need to know the change in volume of the system. The change in volume can be determined by considering the change in the number of moles of gas involved in the reaction.
Since water is being electrolyzed, it is being split into hydrogen and oxygen gases. The balanced chemical equation for the electrolysis of water is:
2 H2O(l) → 2 H2(g) + O2(g)
From the balanced equation, we can see that for every 2 moles of water electrolyzed, 2 moles of hydrogen and 1 mole of oxygen gas are produced.
Given that the mass of water being electrolyzed is 50 g, we can determine the number of moles of water electrolyzed using the molar mass of water. The molar mass of water (H2O) = atomic mass of hydrogen (1.008 g/mol) * 2 + atomic mass of oxygen (16.00 g/mol) = 18.02 g/mol.
Number of moles of water (n) = mass of water / molar mass of water
= 50 g / 18.02 g/mol
≈ 2.7766 mol
Since two moles of water produce two moles of hydrogen gas, the number of moles of hydrogen gas produced (nH2) is also 2.7766 mol.
The change in volume (ΔV) can be calculated using the ideal gas law equation:
ΔV = n * R * ΔT / P
Where:
ΔV = Change in volume
n = Number of moles of gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
ΔT = Change in temperature (in Kelvin)
P = Pressure (in atm)
In this case, the process is performed at constant pressure, so there is no change in pressure. Therefore, the expansion work done is equal to the change in volume (ΔV).
The given temperature is 25°C. To convert this temperature to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
= 25°C + 273.15
= 298.15 K
Now, we can substitute the values into the equation:
ΔV = nH2 * R * ΔT / P
= 2.7766 mol * 0.0821 L·atm/(mol·K) * 298.15 K / 1 atm
≈ 0.675 L·atm
Therefore, the expansion work done when 50 g of water is electrolyzed under constant pressure at 25°C is approximately 0.675 L·atm.