A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O...

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0 L vessel at 300 K. At equilibrium [NO] = 0.062 M.

2 NO(g) + 2H2 ⇌ N2(g) + 2 H2O(g)

a. What is the Kc at 300 K?

b. What is the Kc of the halved reaction, given, at 300 K?

NO(g) + H2 ⇌ ½ N2(g) + H2O(g)

c. What is the Kp of part (a) at 300 K?

d. What is the Kc of the reaction if reversed?

Expert Solution

queen_honey_blossom answered 2 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O...

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2NO(g)+2H2(g)←−→N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentration of H2. Calculate the equilibrium concentration of N2. Calculate the equilibrium concentration of H2O. Calculate Kc

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10. A mixture of 0.08729 mol of H2O, 0.1041 mol of CH4, 0.08069 mol of CO,...

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Question

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O...

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