For the reaction A(g) + B(g) C(g) + D(g), the equilibrium amounts in a 1 L vessel are 4, 4, 2 and 8 moles respectively. If 18 moles of D are now added to the same 1 L container, what is the new equilibrium concentration of C (in M)?

Question

For the reaction A(g) + B(g) <=> C(g) + D(g), the equilibrium amounts in a 1 L vessel are 4, 4, 2 and 8 moles respectively. If 18 moles of D are now added to the same 1 L container, what is the new equilibrium concentration of C (in M)?

Accepted Answer

1 M. Kc = (2*8)/(4*4) = 1. Adding D drives the reverse reaction by x = 1 mol, so C becomes 2 - 1 = 1 mol in 1 L, i.e. 1 M.

For the reaction A(g) + B(g) <=> C(g) + D(g), the equilibrium amounts in a 1 L vessel are 4, 4, 2 and 8 moles respectively. If 18 moles of D are now added to the same 1 L container, what is the new equilibrium concentration of C (in M)?

Solution

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