19.1-21. effect of a slow reaction rate on diffusion. gas a diffuses from point 1 to a catalyst surface at point 2, where it reacts as follows: 2a → b. gas b diffuses back a distance δ to point 1. derive the equation for na for a very fast reaction using mole fraction units xa1, and so on. for dab = 0.2 × 10–4m2/s, xa1 = 0.97, p = 101.32 kpa, δ = 1.30 mm, and t = 298 k, solve for na. do the same as in part (a) but for a slow first-order reaction where k1′ is the reaction velocity constant. calculate na and xa2 for part (c) where k1′=0.53×10−2m/s.

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

listen, you cant how young you were.

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