For the ammonia synthesis reaction, N2(8) +H2 → NH3(g) the equilibrium conversion to ammonia is large at 300 K, but it decreases rapidly with increasing T. However, reaction rates become appreciable...


For the ammonia synthesis reaction,<br>N2(8) +H2 → NH3(g)<br>the equilibrium conversion to ammonia is large at 300 K, but it decreases rapidly with increasing<br>T. However, reaction rates become appreciable only at higher temperatures. For a feed mixture<br>of hydrogen and nitrogen in the stoichiometric proportions,<br>(a) What is the equilibrium mole fraction of ammonia at 1 bar and 300 K?<br>(b) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure<br>of 1 bar?<br>(c) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure<br>of 100 bar, assuming the equilibrium mixture is an ideal gas?<br>(d) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure<br>of 100 bar, assuming the equilibrium mixture is an ideal solution of gases?<br>

Extracted text: For the ammonia synthesis reaction, N2(8) +H2 → NH3(g) the equilibrium conversion to ammonia is large at 300 K, but it decreases rapidly with increasing T. However, reaction rates become appreciable only at higher temperatures. For a feed mixture of hydrogen and nitrogen in the stoichiometric proportions, (a) What is the equilibrium mole fraction of ammonia at 1 bar and 300 K? (b) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 1 bar? (c) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal gas? (d) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal solution of gases?

Jun 08, 2022
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