# Given that the equilibrium...4NH_3(g) + 3O_2(g)rarr2N_2(g) + 6H_2O(l) + Delta is exothermic...what will be effect on the equilibrium if... i. "Oxygen is removed;" ii. "Nitrogen is added;" iii. "Water is removed;" iv. "If heat is added?"

Aug 15, 2015

Apply Le Chatelier's principle, and induce result.

#### Explanation:

For the reaction:

$4 N {H}_{3} \left(g\right) + 3 {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 {N}_{2} \left(g\right) + 6 {H}_{2} O \left(g\right) + h e a t$,

Le Chatelier's principle states that an equilibrium under stress will move in a direction so as to oppose (not counteract!) the external perturbation. All products and reactants are (reasonably!) assumed to be gaseous.

i. Removal of ${O}_{2}$ reactant will drive equilibrium to the LHS; therefore, $\left[N {H}_{3}\right]$ should increase.

ii. Addition of ${N}_{2}$ product will likewise drive equilibrium to LHS; $\left[N {H}_{3}\right]$ should increase.

iii. Addition of ${H}_{2} O$ product will likewise drive equilibrium to LHS; $\left[N {H}_{3}\right]$ should increase.

iv. If temperature is increased, we treat "heat" as a product. (Heat is a virtual product because it is a given that the reaction as written is strongly exothermic.) Equilibrium should be driven to the left, and $\left[N {H}_{3}\right]$ should increase.

v. Should volume of container be increased, concentration (read pressure) of all species (products and reactants) is decreased. Equilibrium will so move as to oppose the change in concentration, and move to the right (because the product side has 8 mol of gas, whereas the reactant side has only 7.)

This is no longer part of the A level syllabus (but is part of 1st year university chemistry), but what do you think would be the effect be of adding an inert gas to the reaction mix; how would the equilibrium evolve?