The value of kc for the following reaction is 0.630 at 409 K N2O4(g) --> 2NO2(g) if a reaction vessel at that temperature intitially contains 0.0250 M NO2 and 0.0250 M N2O4, what is the concentration of NO2 at equilibrium

The concentration of nitrogen dioxide at equilibrium is 0.063 M

Explanation:

We are given:

Initial concentration of nitrogen dioxide = 0.0250 M

Initial concentration of dinitrogen tetraoxide = 0.0250 M

For the given chemical equation:

Initial:                0.025        0.025

At eqllm:         0.025-x     0.025+2x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x', because concentration cannot be negative

So, equilibrium concentration of nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M

Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M

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 !