7.97 mol NH 3 are placed in a 9.7 L flask and allowed to equilibrate at some tem

Question

7.97 mol NH3 are placed in a 9.7 L flask and allowed to equilibrate at some temperature. At equilbrium, 4.47 mol NH3 remain. Fill in the following reaction table in moles and determine the value of Kc at this temperature. Hint: the c in Kc stands for concentration.

2NH3(g) 3H2(g) + N2(g) initial mol mol final mol 7.97 mol NH3 are placed in a 9.7 L flask and allowed to equilibrate at some temperature. At equilbrium, 4.47 mol NH3 remain. Fill in the following reaction table in moles and determine the value of Kc at this temperature. Hint: the c in Kc stands for concentration.

Explanation / Answer

moles of H2(g) formed= 3*3.5/2=5.25moles

mole of N2(g) formed = 1.75 moles

hence Kc = (conc. of H2(g))^3 * (conc of N2(g))/(conc of) NH3(g)^2

Kc = 0.1346 mole^2/liter^2

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