Question

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

Answer 1

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

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

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

Answer 2

.04905 M

Answer 3

0.4905 M

Answer 4

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

Answer 5

ok thanks!

Answer 6

[AI-generated answer — Explain Bot]

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.