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Writing the Reaction Quotient

In Chapter 16, you saw that the rate law for an overall reaction cannot be written from the balanced equation, but must be determined from rate data. In contrast, the reaction quotient can be written directly from the balanced equation Q is a ratio made up of product concentration terms multiplied together and divided by reactant concentration terms multiplied together, with each term raised to the power of its stoichiometric coefficient in the balanced equation. [Pg.545]

The most common form of the reaction quotient shows reactant and product terms as molar concentrations, which are designated by square brackets, [ ]. In the cases you ve seen so far, K is the equilibrium constant based on concentrations, designated from now on as K. Similarly, we designate the reaction quotient based on concentrations as Q - For the general balanced equation [Pg.545]

To construct the reaction quotient for any reaction, write the balanced equation first. For the formation of ammonia from its elements, for example, the balanced equation (with colored coefficients for easy reference) is [Pg.545]

To construct the reaction quotient, we place the product term in the numerator and the reactant terms in the denominator, multiplied by each other, and raise each term to the power of its balancing coefficient (colored as in the equation)  [Pg.545]

SAMPLE PROBLEM 17.1 Writing the Reaction Quotient from the [Pg.545]


Provided that the pressure of hydrogen is 1 bar, we can write the reaction quotient as Q = [H "]2[C1 ]2. To find the concentration of hydrogen ions, we write the Nernst equation ... [Pg.629]

Explain why the following equilibria are heterogeneous and write the reaction quotient Q for each one. [Pg.586]

Problem Write the reaction quotient, Q, for each of the following reactions ... [Pg.545]

Based on the rules for writing the reaction quotient, we have... [Pg.548]

At equilibrium, analysis shows that the flask contains 1.80 mol of H2, 1.80 mol of I2, and 0.520 mol of HI. We calculate Kc by finding the concentrations and substituting them into the reaction quotient. Given the balanced equation, we then write the reaction quotient ... [Pg.552]

Plan First, we use the balanced equation to write the reaction quotient. We can calculate the equilibrium concentrations from the given numbers of moles and the flask volume (0.32 L). Substituting these into 2c and setting it equal to the given K. (0.26), we solve for the unknown equilibrium concentration, [H2O]. [Pg.555]

Plan We have to And the composition of the equilibrium mixture, in other words, the equilibrium concentrations. As always, we use the balanced equation to write the reaction quotient. We find the initial [CO] and [H2O] from the given amounts (0.250 mol of each) and volume (0.125 L), use the balanced equation to define x and set up a reaction table, substitute into 2o and solve for x, from which we calculate the concentrations. [Pg.555]

Plan We write the reaction quotient to see how is affected by each disturbance, relative to K. This effect tells us the direction in which the reaction proceeds for the system to reattain equilibrium and how each concentration changes. [Pg.563]

Changing Value of the Reaction Quotient Writing the Reaction Quotient... [Pg.542]

Writing the Reaction Quotient and Finding K for an Overall Reaction... [Pg.548]

Solution Writing the reaction quotient and using the data to find K ... [Pg.553]


See other pages where Writing the Reaction Quotient is mentioned: [Pg.508]    [Pg.508]    [Pg.586]    [Pg.586]    [Pg.540]    [Pg.545]    [Pg.551]    [Pg.555]    [Pg.555]    [Pg.558]    [Pg.560]    [Pg.563]    [Pg.574]    [Pg.540]    [Pg.545]    [Pg.551]    [Pg.555]    [Pg.555]    [Pg.558]    [Pg.560]    [Pg.563]    [Pg.574]    [Pg.546]    [Pg.548]    [Pg.553]   


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