Suppose a 500. mL flask is filled with 1.6 mol of N, and 0.10 mol of O,. The following reaction becomes possible:N,(g) +0,(g) 2NO(g)The equilibrium constant K for this reaction is 2.74 at the temperature of the flask.Calculate the equilibrium molarity of N,. Round your answer to two decimal places.2:

Answered: Suppose a 500. mL flask is filled with…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Suppose a 500. mL flask is filled with 1.6 mol of N, and 0.10 mol of O,. The following reaction becomes possible:
N,(g) +0,(g) 2NO(g)
The equilibrium constant K for this reaction is 2.74 at the temperature of the flask.
Calculate the equilibrium molarity of N,. Round your answer to two decimal places.
2:

Expert Solution

Step 1

Equilibrium constant $(K_{C})$ for a reaction is a constant and it is the ratio between equilibrium concentration of products and reactants.

The given balanced equation is as follows.

$N_{2 (g)} + O_{2 (g)} \Leftrightarrow 2 {NO}_{(g)}$

$K_{C} = \frac{{[NO]}^{2}}{[N_{2}] [O_{2}]}$

Equilibrium constant for this reaction is $2.74$ .

Step 2

Initial concentration of $N_{2} and O_{2}$ can be calculated as follows.

$\begin{array}{rcl} Concentration & = & \frac{no of moles}{Volume in liters} \\ C_{N_{2}} & = & \frac{1.6 mol}{0.5 L} = 3.2 mol / L = 3.2 M \\ C_{O_{2}} & = & \frac{0.10 mol}{0.5 L} = 0.2 mol / L = 0.2 M \end{array}$

Now we can construct the ICE table.

$N_{2 (g)} + O_{2 (g)} \Leftrightarrow 2 {NO}_{(g)} Initial conc . 3.2 M 0.2 M 0 Change in conc . - x - x + 2 x Equilibrium conc . (3.2 - x) M (0.2 - x) M 2 x$

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