The enthalpy of formation of AB2(l) from A2(g) and B2(g) at 545.5*K is -642.4 kJ/mol. Determine (delta)E for this process at 543.5*K (4 significant figures)
A2(g) + 2 B2 (g) ---------------------> 2 AB2 (l) , H = -642.4 kJ / mol
n = products - reactants
= 0 -(1 + 2 )
= -3
H = E + n RT
-642.4 = E - 3 x 8.314 x 10^-3 x 545.5
E = -628.8 kJ / mol
Get Answers For Free
Most questions answered within 1 hours.