N2+2H2=N2H4 What is the concentration of H2 at equalibrium when (N2)=.033 and (N2H2)= .042 M? Keq=5.29

..........N2 + 2H2 ==> N2H4

aquil..0.033....x......0.042

Keq = (N2H2)/(N2)(H2)^2
Substitute and solve for x.

.04905 M

0.4905 M

Yes but if your prof is picky about the number of significant figures s/he may not give you full credit for that answer.

ok thanks!

To find the concentration of H2 at equilibrium, we can use the equilibrium constant (Keq) expression for the given reaction:

Keq = [N2H4] / ([N2]^2 * [H2]^2)

In this case, we have the values for [N2] and [N2H4], and we need to find [H2].

But before calculating the concentration of H2, we need to rearrange the equation to isolate [H2]:

[H2]^2 = [N2H4] / (Keq * [N2]^2)

Now, substitute the given values into the equation:

[H2]^2 = 0.042 / (5.29 * 0.033^2)

Simplifying:

[H2]^2 = 0.042 / (5.29 * 0.001089)

[H2]^2 = 7.717

Taking the square root of both sides:

[H2] = √(7.717)

[H2] ≈ 2.777 M

Therefore, the concentration of H2 at equilibrium is approximately 2.777 M.