At equilibrium, the concentrations in this system were found to be [n2]=[o2]=0.200 m and [no]=0.600 m. n2(g)+o2(g)↽−−⇀2no(g) if more no is added, bringing its concentration to 0.900 m, what will the final concentration of no be after equilibrium is re‑established?
At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:
Now, the new equilibrium condition, taking into account the change x, becomes:
Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:
Thus, we arrange the equation as:
Finally, the new concentration is:
? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being
the balanced equation for reaction of ca metal with o2 to produce cao is as below
2 ca+ o2→ 2 cao2
the reaction is balanced since the number of atoms in reactant side is equal to number of atoms in product side. for example they are 2 atom of ca in reactant side and 2 atoms of ca in the product side.
the process is a meet the requirement of redox reaction because : calcium is oxidized from oxidation state 0 to 2+ while oxygen is reduced from oxidation state 0 to 2-