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5. Thermodynamics and thermochemistry — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry5. Thermodynamics and thermochemistry1 Mark
MCQ
For $N_2 + 3H_2 \rightarrow  2NH_3, \Delta H = -22\  kcal$, and $E_a = 70\  kcal$. Hence $E_a$ for $2NH_3   \rightarrow N_2+3H_2$ is.....$kcal$
  • ✓
    $92$
  • B
    $70$
  • C
    $48$
  • D
    $22$
✓

Answer

Correct option: A.
$92$
a
$\Delta H =\left(E_a\right)_f-\left(E_a\right)_b$

$-22 =70-\left(E_a\right)_b$

since released the energy

$\left(E_a\right)_b=70+22=92 \,k\, cal$

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