At 1173 K, Keq = 0.0108 for the following reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) The reaction takes place in a 10.0 L vessel at 11
73 K. If a mixture of 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2 is allowed to approach equilibrium, what will happen to the amount of CaCO3? Group of answer choices - It will remain the same
- It will increase
- Not enough information is provided to answer this question
- It will decrease
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
K>Q , reaction will move forward by making more product.
K<Q , reaction will move backward by making more reactant.
Equilibrium constant of the reaction =
Concentration of
Concentration of
Concentration of
This means that equilibrium will move in backward direction by which amount of calcium carbonate will increase.
We are given here that the volume ( V ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V ) = 1.50 L,
Initial moles ( n ) = 3.00 mol,
Final Volume ( n ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V ).
PV / nT = PV / nT - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V / n = V / n - isolate V,
V = V n / n = 1.50 L 2.5 mol / 3.00 mol = ( 1.5 2.5 / 3 ) L = 1.25 L