Consider the problem below: (Equation 1) 4 AICI3 (5) + 2 Al(s) + 3 Cl2(g) - 3 Al,Cla(g) AH = -1051 kJ/mol (Equation 2) 2 Als) + 6 Cl(g) 2 AICI3(5) --- AH = -705.6 kJ/mol (Equation 3) 12 Clg) ---> 6 Cl2g) AH = 1460.4 kJ/mol (Equation 4) 6 AlG) + 9 Cl2(g) -- 3 Al,Cla(g) AH = ??????? The AH for equation 4 is: _(express in kJ)

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