To solve this problem, we need to use the formula for the equilibrium constant (K) which is given by the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

The reaction is: A(g) + B(g) ⇄ 2C(g)

At the start, we have 1.0 mol of A and 3.5 mol of B in a 1 L container. So, the initial concentrations are:

[A] = 1.0 M
[B] = 3.5 M
[C] = 0 M (since the reaction hasn't started yet)

At equilibrium, the concentration of A is 0.5 M. Since the reaction consumes A and B in a 1:1 ratio to produce C, the change in concentration of A (which is 0.5 M) will be the same for B and twice as much for C (since 2 moles of C are produced for every mole of A or B consumed).

So, the equilibrium concentrations are:

[A] = 1.0 M - 0.5 M = 0.5 M
[B] = 3.5 M - 0.5 M = 3.0 M
[C] = 0 M + 2*0.5 M = 1.0 M

Now we can substitute these values into the formula for K:

K = [C]^2 / ([A]*[B])
K = (1.0 M)^2 / (0.5 M * 3.0 M)
K = 1.0 / 1.5 = 0.67 (to 2 decimal places)

Question

To solve this problem, we need to use the formula for the equilibrium constant (K) which is given by the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

The reaction is: A(g) + B(g) ⇄ 2C(g)

At the start, we have 1.0 mol of A and 3.5 mol of B in a 1 L container. So, the initial concentrations are:

[A] = 1.0 M
[B] = 3.5 M
[C] = 0 M (since the reaction hasn't started yet)

At equilibrium, the concentration of A is 0.5 M. Since the reaction consumes A and B in a 1:1 ratio to produce C, the change in concentration of A (which is 0.5 M) will be the same for B and twice as much for C (since 2 moles of C are produced for every mole of A or B consumed).

So, the equilibrium concentrations are:

[A] = 1.0 M - 0.5 M = 0.5 M
[B] = 3.5 M - 0.5 M = 3.0 M
[C] = 0 M + 2*0.5 M = 1.0 M

Now we can substitute these values into the formula for K:

K = [C]^2 / ([A]*[B])
K = (1.0 M)^2 / (0.5 M * 3.0 M)
K = 1.0 / 1.5 = 0.67 (to 2 decimal places)

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for the equilibrium constant (K) which is given by the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

The reaction is: A(g) + B(g) ⇄ 2C(g)

At the start, we have 1.0 mol of A and 3.5 mol of B in a 1 L container. So, the initial concentrations are:

[A] = 1.0 M
[B] = 3.5 M
[C] = 0 M (since the reaction hasn't started yet)

At equilibrium, the concentration of A is 0.5 M. Since the reaction consumes A and B in a 1:1 ratio to produce C, the change in concentration of A (which is 0.5 M) will be the same for B and twice as much for C (since 2 moles of C are produced for every mole of A or B consumed).

So, the equilibrium concentrations are:

[A] = 1.0 M - 0.5 M = 0.5 M
[B] = 3.5 M - 0.5 M = 3.0 M
[C] = 0 M + 2*0.5 M = 1.0 M

Now we can substitute these values into the formula for K:

K = [C]^2 / ([A]*[B])
K = (1.0 M)^2 / (0.5 M * 3.0 M)
K = 1.0 / 1.5 = 0.67 (to 2 decimal places)

Samples of A (1.0 mol) and B (3.5 mol) are placed in a 1 L container and the following reaction takes place:A(g) + B(g) ⇄ 2C(g) At equilibrium, the concentration of A is 0.5 M. What is the value of K to 2 d.p?

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