\begin{aligned} & \frac{1}{2} A \longrightarrow B ; \quad \Delta H=150 \mathrm{~kJ} / \mathrm{mol} \\ & 3 B \longrightarrow 2 C+D ; \Delta H=-125 \mathrm{~kJ} / \mathrm{mol} \\ & E+A \longrightarrow 2 D ; \Delta H=+350 \mathrm{~kJ} / \mathrm{mol} \\ & \hline \text { By }[2 \times(\mathrm{i})+(\mathrm{ii})]-(\mathrm{iii}), \text { we have } \\ & B+D \longrightarrow E+2 C \\ & \therefore \quad \Delta H=150 \times 2+(-125)-350 \\ & \quad=-175 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}
Consider the following processes Δ H( ~kJ / mol) … For B+D → E+2 C, Δ H will be
NEET UG 2011 — Chemistry Physical Chemistry
Consider the following processes ΔH( kJ/mol) 1/2A→B3B→2C+DE+A→2D+150−125+350 For B+D→E+2C,ΔH will be
Held on 30 Apr 2011 · Verified 9 Jul 2026.
Options
- A
525 kJ/mol
- B
−175 kJ/mol
- C
−325 kJ/mol
- D
325 kJ/mol
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