3A (g) ⇆ B (g) + C (g) + D (g), K_c for the following reactions is 27. If number of moles of B at equilibrium are 3 then calculate number of initial moles of A taken in the container ?
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Let N moles of A be present initially in the container of fixed volume V litres. 3 moles of B , C and D each are produced at the equilibrium.
To produce 3 moles of B, nine moles of A are consumed, as 3 moles of A are needed for 1 mole of B. Also 3 moles each of D & C are produced. N-9 moles of A remain.
Total number of moles of gases at the equilibrium is N-9+3+3+3 = N+8.
3 A (g) ↔ B (g) + C (g) + D (g)
N-9 moles 3 moles 3 moles 3 moles
[ B ] = 3/V = [ C ] = [ D ] and [ A ] = (N-9)/V
Kc = 27 = [ B ] [ C ] [ D ] / [ A ]³
= 3³ / (N-9)³
Solving that we get N = 10. Answer.
To produce 3 moles of B, nine moles of A are consumed, as 3 moles of A are needed for 1 mole of B. Also 3 moles each of D & C are produced. N-9 moles of A remain.
Total number of moles of gases at the equilibrium is N-9+3+3+3 = N+8.
3 A (g) ↔ B (g) + C (g) + D (g)
N-9 moles 3 moles 3 moles 3 moles
[ B ] = 3/V = [ C ] = [ D ] and [ A ] = (N-9)/V
Kc = 27 = [ B ] [ C ] [ D ] / [ A ]³
= 3³ / (N-9)³
Solving that we get N = 10. Answer.
kvnmurty:
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