Question

At a particular temperature, \(K=3.75\) for the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ If all four gases had initial concentrations of \(0.800 M\), calculate the equilibrium concentrations of the gases.

Short answer

The equilibrium concentrations for the given reaction are: \(\mathrm{SO}_{2}\): \(0.545M\), \(\mathrm{NO}_{2}\): \(0.545M\), \(\mathrm{SO}_{3}\): \(1.055M\), and \(\mathrm{NO}\): \(1.055M\).

Step-by-step solution

Write the equilibrium expression

To start, we need to write an equilibrium expression for the given chemical reaction. The equilibrium expression is the ratio of the product concentrations to the reactant concentrations. Each concentration is raised to the power of its stoichiometric coefficient. In this case, the equilibrium expression is: $$ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} $$

Set up the ICE table

An ICE table (Initial, Change, Equilibrium) helps to organize the information in a systematic way. For this problem, the ICE table would look like: | | Initial | Change | Equilibrium | |:-:|---------|--------|-------------| | SO2 | 0.800 | -x | 0.800-x | | NO2 | 0.800 | -x | 0.800-x | | SO3 | 0.800 | +x | 0.800+x | | NO | 0.800 | +x | 0.800+x | The change in the concentrations is represented by "x". The reactants' concentrations decrease by x, while the products' concentrations increase by x.

Substitute equilibrium concentrations into the equilibrium expression

We now substitute the equilibrium concentrations into the equilibrium expression we derived in Step 1: $$ 3.75 = \frac{(0.800+x)(0.800+x)}{(0.800-x)(0.800-x)} $$

Solve for x

Next, we solve for x using quadratic equation or other algebraic methods. For the given equilibrium constant, the equation simplifies as: $$ 3.75 = \frac{(0.800+x)^{2}}{(0.800-x)^{2}} $$ Now, take the square root of both sides: $$ \sqrt{3.75} = \frac{0.800+x}{0.800-x} $$ $$ 1.935 = \frac{0.800+x}{0.800-x} $$ We can now cross multiply and solve for x: $$ 1.935(0.800-x)=(0.800+x) $$ $$ 1.548 - 1.935x = 0.800+x $$ $$ 2.935x = 0.748 $$ $$ x = 0.255 $$

Calculate the equilibrium concentrations

Finally, we can plug in x to calculate the equilibrium concentrations: $$ [\mathrm{SO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{NO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{SO}_{3}] = 0.800 + 0.255 = 1.055 M $$ $$ [\mathrm{NO}] = 0.800 + 0.255 = 1.055 M $$ Now, we have the equilibrium concentrations for all gases: SO2: 0.545 M NO2: 0.545 M SO3: 1.055 M NO: 1.055 M