A chemist is studying the rate of the Haber synthesis: N2 + 3H2 2NH3 Starting with a closed reactor containing 1.25 mol/L of N2 and 0.50 mol/L of H2, the chemist finds that the H2 concentration has fallen to 0.25 mol/L in 44 seconds. What is the N2 concentration after 44 seconds?

1.17 M

Explanation:

Step 1: Write the balanced equation

N₂ + 3 H₂ ⇒ 2 NH₃

Step 2: Calculate the rate of disappearance of H₂

We will use the following expression.

rH₂ = - Δ[H₂]/t = - (0.25 M - 0.50 M)/44 s = 0.0057 M/s

Step 3: Calculate the rate of disappearance of N₂

The molar ratio of N₂ to H₂ is 1:3.

0.0057 mol H₂/L.s × 1 mol N₂/3 mol H₂ = 0.0019 mol N₂/L.s

Step 4: Calculate the final concentration of N₂

We will use the following expression.

[N₂] = [N₂]₀ - rN₂ × t

[N₂] = 1.25 mol/L - 0.0019 mol/L.s × 44 s

[N₂] = 1.17 M

answer is d because it is trust me

the answer is definitely c.