Question

15 moles of N2 is mixed with 20 moles of H2 in an 8 L vessel. 5.6 moles of ammonia is formed. Calculate Kc for the equation; N2(g) + 3H2(g) <---> NH3(g) + heat

Answer 1

Answer:

based upon the formula of equilibrium constant formula and please make it as brilliest answer

Answer 2

hence, value of equilibrium constant would be 0.016 L²/mol²

reaction is given....

N₂(g) + 3H₂(g) ⇒2NH₃(g)

at t = 0. 15 20. 0

at eqlm 15 - x 20 - 3x 2x

given, 2x = 5.6 ⇒x = 2.8

so, 15 - x = 15 - 2.8 = 12.2

20 - 3x = 20 - 3 × 2.8 = 20 - 8.4 = 11.6

now, [NH₃] = 5.6/8 = 0.7

[N₂] = 12.2/8 = 1.525

[H₂] = 11.6/8 = 1.45

now K_c = [NH₃]²/[N₂][H₂]³

= (0.7)²/(1.525)(1.45)³

= 0.049/(1.525 × 2.0125)

= 0.016

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