Consider the system 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) ∆H = -1530.4 kJ How will the amount of ammonia (NH3) at equilibrium be affected by (a) expanding the container at constant temperature? (b) increasing the temperature?
Added by David L.
Step 1
According to Le Chatelier's principle, the system will shift in the direction that will increase the pressure. In this case, the reaction will shift towards the side with more moles of gas. The reactant side has 4 moles of NH3 and 3 moles of O2, totaling 7 moles Show more…
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For the equilibrium system: 4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) + 906 kJ (1) How does an increase in temperature affect [NH3]? (2) How does an increase in the volume of the container affect [NH3]?
Madhur L.
Consider the system $4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-1530.4 \mathrm{~kJ}$ (a) How will the amount of ammonia at equilibrium be affected by 1. removing $\mathrm{O}_{2}(g) ?$ 2. adding $\mathrm{N}_{2}(g) ?$ 3. adding water? 4. expanding the container at constant pressure? 5. increasing the temperature? (b) Which of the above factors will increase the value of $K ?$ Which will decrease it?
Consider the system $$\begin{array}{r}4 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \\\Delta H=-1530.4 \mathrm{~kJ}\end{array}$$ How will the amount of ammonia at equilibrium be affected by (1) removing $\mathrm{O}_{2}(\mathrm{~g})$ ? (2) adding $\mathrm{N}_{2}(\mathrm{~g})$ ? (3) adding water? (4) expanding the container at constant pressure? (5) increasing the temperature?
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