For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} M\), \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Short Answer

Expert verified
Given the equilibrium concentrations \([\mathrm{NO}(g)] = 8.1 \times 10^{-3}\,M\), \([\mathrm{H}_{2}(g)] = 4.1 \times 10^{-5}\,M\), \([\mathrm{N}_{2}(g)] = 5.3 \times 10^{-2}\,M\), and \([\mathrm{H}_{2}\mathrm{O}(g)] = 2.9 \times 10^{-3}\,M\), we can calculate the equilibrium constant \(K\) using the expression \(K = \frac{[\mathrm{N}_{2}] \times [\mathrm{H}_{2} \mathrm{O}]^2}{[\mathrm{NO}]^2 \times [\mathrm{H}_{2}]^2}\). Substituting the given concentrations, we get \(K \approx 13.09\) for the reaction at this temperature.

Step by step solution

01

Write down the given concentrations

The concentrations at equilibrium are given as: $$ [\mathrm{NO}(g)] = 8.1 \times 10^{-3}\,M, [\mathrm{H}_{2}(g)] = 4.1 \times 10^{-5}\,M, [\mathrm{N}_{2}(g)] = 5.3 \times 10^{-2}\,M, \text{ and } [\mathrm{H}_{2}\mathrm{O}(g)] = 2.9 \times 10^{-3}\,M $$
02

Substitute the concentrations in the equilibrium constant expression.

Using the equilibrium constant expression, replace the concentrations: $$ K = \frac{(5.3 \times 10^{-2}) \times (2.9 \times 10^{-3})^2}{(8.1 \times 10^{-3})^2 \times (4.1 \times 10^{-5})^2} $$
03

Calculate the value of K.

Compute the value of the equilibrium constant: $$ K = \frac{(5.3 \times 10^{-2}) \times (2.9 \times 10^{-3})^2}{(8.1 \times 10^{-3})^2 \times (4.1 \times 10^{-5})^2} \approx 13.09 $$ So, the equilibrium constant \(K\) for the reaction at this temperature is approximately \(13.09\).

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

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.

A 1.00-L flask was filled with \(2.00\) mol gaseous \(\mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol}\) gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) mol gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Old-fashioned "smelling salts" consist of ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammonium carbonate $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

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