Chapter 6: Q5P (page 141)

For the reaction $2 A (g) + B (a q) + 3 C (l) ⇌ D (s) + 3 E (g)$ , the concentrations at equilibrium are found to be
A: $2.8 \times 10^{3} Pa$
B: $1.2 \times 10^{- 2} M$
C: $12.8 M$
D: $16.5 M$
E: $3.6 \times 10^{4} Torr$
Find the numerical value of the equilibrium constant that would appear in a conventional table of equilibrium constants.

Short Answer

Expert verified

The numerical value of the equilibrium constant is $1.2 \times 10^{10}$ .

Step by step solution

The equilibrium constant.

Equilibrium constant (K): In the equilibrium reaction, the ratio of the concentration of the reactant and the concentration of the product is taken. If K's value is smaller than 1, the reaction should be moved to the left; if K is more than 1, the reaction should be moved to the right.

Example of an equilibrium reaction:

$a A + b B \overset{K}{⇌} c C + d D$

$K = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {| B |}^{b}}$

The numerical value of the equilibrium constant for the given reaction.

The given equation is,

$2 A_{(g)} + B_{(a q)} + 3 C_{(I)} ⇌ D_{(s)} + 3 E_{(g}$

$A = 2.8 \times 10^{3} Pa B = 1.2 \times 10^{- 2} M C = 12.8 MD = 16.5 M E = 3.6 \times 10^{4} Torr$

The numerical value of the equilibrium constant for the given reaction:

$K = \frac{P_{E}^{3}}{P_{A}^{2} [B]}$

$K = \frac{{(\frac{3.6 \times 10^{4} Torr}{760 T or / atm} \times 1.013 \frac{bar}{atm})}^{3}}{{(\frac{2.8 \times 10^{3} pa}{10^{5} Pabar})}^{2} (1.2 \times 10^{- 2} M)} = 1.2 \times 10^{10}$