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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2009 · Chemistry · Thermodynamics (C)

Which one of the following is correct regarding Δ H:

Which one of the following is correct regarding ΔH:\Delta \mathrm{H}:

Options

  1. AΔH=ΔE+2RT\quad \Delta \mathrm{H}=\Delta \mathrm{E}+2 \mathrm{RT}
  2. BΔH=ΔE−2RT\quad \Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}
  3. CΔH=ΔE+RT\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{RT}
  4. DΔH=ΔE−RT\quad \Delta \mathrm{H}=\Delta \mathrm{E}-\mathrm{RT}

Correct answer

B. ΔH=ΔE−2RT\quad \Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}

Step-by-step solution

N2+3H2⇌2NH3\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} According to thermodynamics's Ist law ΔH=ΔE+nRT\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{nRT} Where ΔH=\Delta \mathrm{H}= enthalpy of reaction at constant pressure ΔE=\Delta \mathrm{E}= heat of reaction at constant volume R=\mathrm{R}= molar gas constant T= temperature of the reaction n=( no. of moles of product )−( no. of moles \begin{array}{l}\mathrm{T}=\text{ temperature of the reaction }\\ \mathrm{n}=(\text{ no. of moles of product })-(\text{ no. of moles } \end{array} of reactant.) From reaction, n=np−nR=2−4=−2\mathrm{n}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{R}}=2-4=-2 Hence, ΔH=ΔE−2RT\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}.
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