Consider the dehydrogenation of ethane to form ethene (ethylene) and hydrogen, latex: \text{c}_2\text{h}_6\rightleftharpo ons\text{c}_2\text{h}_4+\text{h}_2c 2 h 6 ⇌ c 2 h 4 + h 2 the standard enthalpies of formation and the standard entropies at 25 °c and 1 atm are given below, as pulled from oxtoby, gillis, and nachtrieb, principles of modern chemistry, fourth edition; fort worth: saunders, 1999. substance latex: \delta h^\circ_fδ h f ∘ (kj/mol) latex: s^\circs ∘ (j/mol/k) carbon (s, graphite) 5.74 hydrogen (g) 130.57 ethylene (g) 52.26 219.45 ethane (g) −84.68 229.49 assuming your goal is the production of hydrogen and/or ethylene, would it be beneficial to carry out this reaction at high pressure or low pressure, assuming kinetics are not an issue (i. e., that equilibrium is achieved reasonably quickly at any pressure)?

It would be better to carry out the reaction at low pressure

Explanation:

[\tex]C_2H_6 (g)\rightleftharpoons C_2H_4 (g)+H_2 (g)[/tex]

Assuming that the reaction above is in equilibrium and getting based on the principles of Le Chatellier we can say that:

If we increase the pressure in the system, the reaction will go Leftward because it will try to reduce the pressure by consuming moles of gas.

If we reduce the pressure in the system, the reaction will go rightward because it will try to increase the pressure by producing moles of gas.

Given that our goal is to produce ehtylene, we want the reaction to go rightward and to do so is better to work at low pressure.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2