Verified Solution • 2025–26 Session moderate

QAIPM-2011-Ch-1788: NEET 2011 Chemistry - Class 11 AIPM-2011-Ch-1788 Solution: Consider the following processes: \( H\ ( {kJ/mol})\) \(1/2A…

Consider the following processes: ΔH (kJ/mol)\Delta H\ (\text{kJ/mol}) 1/2A→B1/2A\rightarrow B +150; 3B→2C+D3B\rightarrow 2C+D -125; E+A→2DE+A\rightarrow 2D +350. For B+D→E+2CB+D\rightarrow E+2C, ΔH\Delta H will be

Multiple Choice Options

  • A 525 kJ/mol
  • B −175-175 kJ/mol
  • C −325-325 kJ/mol
  • D 325 kJ/mol

Step-by-Step Verified Solution

H (kJ/mol) 1/2A _ B | +150 3B   _ 2C + D | -125 E + A   _ 2D | +350 ___________________________________    B + D _ E + 2C; _ H = (300 - 125 - 350) = - 175 kJ/mol

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