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Chemistry · 2011

NEET 2011 · Mains · Code B · Q14

Consider the following processes: Δ H ( kJ / mol ) 1 / 2 A → B +150 3 B → 2 C + D -125 E + A → 2 D +350 For B + D → E +2 C, Δ H will be:

Consider the following processes : $\displaystyle \Delta \mathrm{H}(\mathrm{kJ} / \mathrm{mol})$ $\displaystyle 1 / 2 \mathrm{~A} \rightarrow \mathrm{~B} \quad+150$ $\displaystyle 3 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D} \quad-125$ $\displaystyle \mathrm{E}+\mathrm{A} \rightarrow 2 \mathrm{D} \quad+350$ For $\displaystyle \mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}+2 \mathrm{C}, \Delta \mathrm{H}$ will be :
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NEET (UG) 2011 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.