DE = dH - PdV

2 H2O(g) → 2 H2(g) + O2(g) 

You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. 

1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. 

Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas 

2 moles reacts to form 3 moles 

The gas equation is 

PV = nRT 
P = pressure 
V = volume (unknown) 
n = moles (1) 
R = gas constant = 8.314 J K^-1 mol^-1 
- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) 
T = temp in Kelvin (kelvin = deg C + 273.15 
So T = 403.15 K 

Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say 

dE = dH -dnRT (because PV = nRT) 

Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules 

dH = 483.6 kJ/mol = 483600 Joules/mol 

dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) 
dE = 483600 J/mol - 3351.77 J 
dE = 480248.23 J/mol 
dE = 480.2 kJ/mol

5 0

3 years ago

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What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?

Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

4 0

3 years ago