Question

What happens if the following reaction is at equilibrium in a sealed container and then the volume is doubled and the temperature is not changed? 2F2(g) + 2BrF(g) ? 2BrF3(l) K=1 × 10^46 ?H°=-413.9 kJ mol?¹ Reactants will form more products and the equilibrium constant will increase. Products will form more reactants and the equilibrium constant will decrease. Reactants will form more products and the equilibrium constant will stay the same. Products will form more reactants and the equilibrium constant will stay the same. Nothing. The system will remain at equilibrium and the equilibrium constant will not change.

          What happens if the following reaction is at equilibrium in a sealed container and then the volume is doubled and the temperature is not changed?
2F2(g) + 2BrF(g) ? 2BrF3(l) K=1 × 10^46 ?H°=-413.9 kJ mol?¹
Reactants will form more products and the equilibrium constant will increase.
Products will form more reactants and the equilibrium constant will decrease.
Reactants will form more products and the equilibrium constant will stay the same.
Products will form more reactants and the equilibrium constant will stay the same.
Nothing. The system will remain at equilibrium and the equilibrium constant will not change.

What happens if the following reaction is at equilibrium in a sealed container and then the volume is doubled and the temperature is not changed?
2F2(g) + 2BrF(g) ? 2BrF3(l) K=1 × 10^46 ?H°=-413.9 kJ mol?¹
Reactants will form more products and the equilibrium constant will increase.
Products will form more reactants and the equilibrium constant will decrease.
Reactants will form more products and the equilibrium constant will stay the same.
Products will form more reactants and the equilibrium constant will stay the same.
Nothing. The system will remain at equilibrium and the equilibrium constant will not change.

Added by Elizabeth H.

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