Calculate the change in internal energy at 1 atm & 25°C.
H2 (g) + 1/2 O2 (g) ---> H2O (l)
delta(H)= -286 KJ/mol
Okay so i started with the equation:
delta(E) = delta(H) - RTdelta(n)
P= 1 atm
T= 298 K
but i don't exactly know what to do from here..please help thank you.
To calculate the change in internal energy (ΔE) at 1 atm and 25°C, you can use the equation:
ΔE = ΔH - RTΔn
Given:
ΔH = -286 kJ/mol (change in enthalpy)
P = 1 atm (pressure)
T = 25°C = 298 K (temperature)
However, before we can use this equation, we need to determine Δn, the change in the number of moles of gas.
From the balanced chemical equation:
H2 (g) + 1/2 O2 (g) → H2O (l)
We can see that the number of moles of gas decreases from 2 to 0.5. Therefore, Δn = 0.5 - 2 = -1.5.
Now we can substitute the values into the equation:
ΔE = -286 kJ/mol - (8.314 J/(mol·K) * 298 K * (-1.5))
= -286 kJ/mol + 3731 J/mol
= -286 kJ/mol + 3.731 kJ/mol
= -282.269 kJ/mol
So, the change in internal energy (ΔE) at 1 atm and 25°C is approximately -282.269 kJ/mol.