For the reaction A(g) + 2B(g) C(g), the equilibrium constant is Kc = 0.25 dm⁶ mol⁻². In a 5 dm³ vessel, how many moles of A must be combined with 4 mol of B so that 1 mol of C is present at equilibrium?

Question

For the reaction A(g) + 2B(g) <=> C(g), the equilibrium constant is Kc = 0.25 dm⁶ mol⁻². In a 5 dm³ vessel, how many moles of A must be combined with 4 mol of B so that 1 mol of C is present at equilibrium?

Accepted Answer

26 moles. At equilibrium B = 4 - 2 = 2 mol, C = 1 mol, A = (a-1) mol. Converting to molarity (divide by 5) and using Kc = 0.25 gives a-1 = 25, so a = 26 mol.

For the reaction A(g) + 2B(g) <=> C(g), the equilibrium constant is Kc = 0.25 dm⁶ mol⁻². In a 5 dm³ vessel, how many moles of A must be combined with 4 mol of B so that 1 mol of C is present at equilibrium?

Solution

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