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?

0.84M

Explanation:

Hello,

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:

Best regards.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !