Q. Consider the following processes: (1/2)A → B; ΔH = +150 kJ/mol 3B → 2C + D; ΔH = −125 kJ/mol E + A → 2D; ΔH = +350 kJ/mol For B + D → E + 2C, ΔH will be
A) −325 kJ/mol
B) 325 kJ/mol
C) 525 kJ/mol
D) −175 kJ/mol
Explanation
[Solution from source] Double the first equation (A → 2B, +300), add the second (3B → 2C + D, −125) and reverse the third (2D → E + A, −350). Adding gives B + D → E + 2C with ΔH = 300 − 125 − 350 = −175 kJ/mol. Forgetting to double the first equation leads to −325 kJ/mol. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.