Mol NH_3 are placed in a 10.2 L flask and allowed to equilibrate at some temperature....

Question

Question

Mol NH_3 are placed in a 10.2 L flask and allowed

science chemistry

Answer 1

Answer #1

2NH3(g) --------> 3H2(g) + N2(g)

I 3.36 0 0

C -2*1.33 3*1.33 1.33

E 0.7 3.99 1.33

Kc = [N2][H2]³/[NH3]²

[NH3] = 0.7/10.2 = 0.068M

[H2] = 3.99/10.2 = 0.39M

[N2] = 1.33/10.2 = 0.13M

Kc = [N2][H2]³/[NH3]²

= 0.13*(0.39)³/(0.068)²

= 0.019773/0.004628 = 4.2724

Answer 2

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