1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant Kp for the reaction, N 2+3 H 2 leftharpoons 2 NH 3 is 1.64 × 10-4 at 400° C and 0.144 × 10-4 at 500° C. Then the mean heat of formation of 1 mole of NH 3 from its elements in this temperature range.

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Q. The equilibrium constant $K_{p}$ for the reaction,
$N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}$ is $1.64 \times 10^{-4}$ at $400^{\circ} C$ and $0.144 \times 10^{-4}$ at $500^{\circ} C$. Then the mean heat of formation of $1$ mole of $NH _{3}$ from its elements in this temperature range______.

Equilibrium

Solution:

We know that, $\log \frac{K_{2}}{K_{1}}=\frac{\Delta H}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$

$\log \frac{0.144}{1.64}=\frac{\Delta H }{2.303 \times 1.987 \times 10^{-3}}\left(\frac{1}{673}-\frac{1}{773}\right)$

$\Delta H =-25.14\, kcal$ for $2\, mole$

$=12.57\, kcal\, mol ^{-1}$