The reaction of nitrogen with hydrogen to form ammonia is thermodynamically favorable.
$\mathrm{N}_{2}(\mathrm{~g}) +3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad\quad\quad\quad\quad\quad\quad \Delta H^{\circ}=-92.2 \mathrm{~kJ}$ J
The equilibrium constant for this reaction is 6.0 × 10⁵ at 298 K.At what temperature is the equilibrium constant equal to 1.0 × 10³? (R = 8.31 J/mol⋅K)

Question

The reaction of nitrogen with hydrogen to form ammonia is thermodynamically favorable.

\mathrm{N}_{2}(\mathrm{~g}) +3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad\quad\quad\quad\quad\quad\quad \Delta H^{\circ}=-92.2 \mathrm{~kJ}

J
The equilibrium constant for this reaction is 6.0 × 10⁵ at 298 K.At what temperature is the equilibrium constant equal to 1.0 × 10³? (R = 8.31 J/mol⋅K)

Quizplus · Accepted Answer

The Answer of The reaction of nitrogen with hydrogen to form ammonia is thermodynamically favorable. $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad\quad\quad\quad\quad\quad\quad \Delta H^{\circ}=-92.2 \mathrm{~kJ}$J The equilibrium constant for this reaction is 6.0 × 10⁵ at 298 K.At what temperature is the equilibrium constant equal to 1.0 × 10³? (R = 8.31 J/mol⋅K)

The Reaction of Nitrogen with Hydrogen to Form Ammonia Is