Question

Consider the following equations:

3A + 6B --> 3D ΔH = -403 kJ/mol

E + 2F --> A   ΔH = -105.2 kJ/mol

C --> E + 3D ΔH = +64.8 kJ/mol

Suppose the first equation is reversed and multiplied by 1/6, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction

Answer 1

Answer to the problem: 3A+ 6 B -> 30 E + 2F - A - e E +31 (1) (11) (1) AH, = - 403 kJ/mol 414 = - 105.2 kJ/mor 44 = +64.8 Kg/mol According to the instruction given in the question ie first equation is reversed and multiplied by t, the second and third equations are divided by 2 and the three equation are added - (3D - 34+ 6B) x + (E+28 - Axį lc - E+13))xtra : 30 + XE + ZF 9A + 6/B + A/2 + E/2 + ²0 +% Net equection 2F+C 24+ 2B + 2D Overall Heat of the reaction = (AH2 + 4H₂ ) x 2 - Allax to = (-105.2 +64.8)x3,- (-403)xt = – 20.6 + 67.17 = 46:57 Kg mot