du = dq + dw
du = +483.6 KJ/Mol + 42.7 L * 1*10^5
Consider the reaction
2H2O(g) → 2H2(g) + O2(g) ΔH = +483.60 kJ/mol
at a certain temperature. If the increase in volume is 42.7 L against an external pressure of 1.00 atm, calculate ΔU for this reaction. (The conversion factor is 1 L · atm = 101.3 J.)
2 answers
The above by Cameron is correct EXCEPT the dw portion.
In L*atm that is 42.7 L x 1 atm. Convert that to J by 42.7 x 1 x 101.3 = dW.
In L*atm that is 42.7 L x 1 atm. Convert that to J by 42.7 x 1 x 101.3 = dW.
1 answer