Question

For the nitrogen fixation reaction, 3h2(g) + n2(g) 2nh3(g), kc = 6.0 × 10–2 at 500°c. if 0.250 m h2 and 0.050 m nh3 are present at equilibrium, what is the equilibrium concentration of n2?a. 3.3 m

b. 2.7 m
c. 0.20 m
d. 0.083 m
e. 0.058 m

Answer 1

Write the Kc equation...
Substitute the given values in the equation and u qill ontain the answer... Give it a try

Answer 2

b. 2.7 M

Step-by-step explanation:

Hello,

In this case, by considering the law of mass action for the studied reaction:

`Kc=([NH_3]^2_(eq))/([H_2]^3_(eq)[N_2]_(eq))`

Now, since the concentration
`N_2` is the unique unknown, solving for it one obtains:

`[N_2]_(eq)=([NH_3]^2_(eq))/([H_2]^3_(eq)Kc)`

`[N_2]_(eq)=((0.050M)^2)/((0.250M)^3(6.0x10^(-2)))`

`[N_2]_(eq)=2.7M`

Hence the answer is b.

Best regards.