Equilibrium 1... 1) The reaction 3A(g) + B(s) 2C (aq) + D (aq) at 25 degree C oc

Question

Equilibrium 1...

1) The reaction 3A(g) + B(s) 2C (aq) + D (aq) at 25 degree C occurs in a 1.87 L flask. After the reaction attains equilibrium, the flask is found to contain 2.48 mol of A, 2.11 mol of C and 4.25 mol of D. Calculate the equilibrium constant for this reaction. ?Keep in mind that you can only calculate the equilibrium constant from equilibrium concentrations and vice-versa. ? So if the problem only has initial (or non-equilibrium) concentrations, you first need to calculate the equilibrium concentrations before finding the equilibrium constant. ? Alternatively you can also use the initial (or non-equilibrium) concentrations and the equilibrium constant to determine equilibrium concentrations. ? We can do this because stoichiometry relates the reactants to the products and we can use this to determine how much products is formed and how much reactant is lost

Explanation / Answer

Kc = [C]^2 [D]/ {[A]^3}
     =2.11^2 * 4.25 /2.48^3
     =1.24

Since numerator and denominator have same overall power we can substitute concentration by number of moles because 1.87 L gets cancelled

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