Question

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature. 56. At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 3.3 × 10−3% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction. 2NO2 (g) ⇌ 2NO(g) + O2 (g)

Answer 1

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

                    $PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial          0.72 mol     0         0

at eq.     0.72 - 0.40   0.40      0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of  $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

                                  $2NO_2 \rightleftharpoons 2NO + O_2$

initial                            1.0 M       0           0

at eq. concentration of  $NO_2$   is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$      

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$