$33.6 \text{ g}$ of $N_2$ and 24 g of $H_2$ are placed in a $5 \text{ dm}^3$ sealed container and react according to the following balanced equation until dynamic equilibrium is reached.
$N_2(g)+3 H_2(g) \rightleftharpoons 2 NH_3(g)$
6.1 Explain the term "dynamic chemical equilibrium"
At 573 K when equilibrium is reached, there is $5.6 \text{ g}$ of $N_2$ gas left inside the container.
6.2 Calculate the concentration of both reactants when the reaction reached equilibrium for the first time.
6.3 Calculate the equilibrium constant $(K_c)$ when this reaction reached equilibrium.
6.4 How would the yield of $NH_3$ change if a smaller container than the $5 \text{ dm}^3$ is used? Write only INCREASES, DECREASES or REMAINS THE SAME.
The temperature of this reaction increases from 573 K to 700 K. At this new temperature the equilibrium constant for this reaction decreases.
6.5 Is the forward reaction endothermic or exothermic?
6.6 Explain your answer to QUESTION 6.5.
Answer (2)
Dynamic chemical equilibrium occurs when the rates of forward and reverse reactions are equal, resulting in constant reactant and product concentrations.
Calculate equilibrium concentrations: [ N 2 ] = 0.04 M and [ H 2 ] = 1.8 M .
Determine the equilibrium constant: K c = 0.686 .
Decreasing the container volume increases the yield of N H 3 .
The forward reaction is exothermic.
Increasing temperature shifts the equilibrium towards reactants, decreasing K c , indicating an exothermic forward reaction. E x o t h er mi c
Explanation
Explanation of Dynamic Chemical Equilibrium Dynamic chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant, but the reactions continue to occur.
Calculating Equilibrium Concentrations First, calculate the initial moles of N 2 and H 2 :
n ( N 2 ) ini t ia l = 28 g / m o l 33.6 g = 1.2 m o l n ( H 2 ) ini t ia l = 2 g / m o l 24 g = 12 m o l Next, calculate the moles of N 2 at equilibrium: n ( N 2 ) e q u i l ib r i u m = 28 g / m o l 5.6 g = 0.2 m o l Then, calculate the moles of N 2 that reacted: n ( N 2 ) re a c t e d = 1.2 m o l − 0.2 m o l = 1.0 m o l Now, calculate the moles of H 2 that reacted: n ( H 2 ) re a c t e d = 3 ∗ n ( N 2 ) re a c t e d = 3 ∗ 1.0 m o l = 3.0 m o l Calculate the moles of H 2 at equilibrium: n ( H 2 ) e q u i l ib r i u m = 12 m o l − 3.0 m o l = 9.0 m o l Finally, calculate the concentrations of N 2 and H 2 at equilibrium: [ N 2 ] = 5 d m 3 0.2 m o l = 0.04 M [ H 2 ] = 5 d m 3 9.0 m o l = 1.8 M
Calculating the Equilibrium Constant To calculate the equilibrium constant K c , we first need to calculate the moles of N H 3 formed: n ( N H 3 ) f or m e d = 2 ∗ n ( N 2 ) re a c t e d = 2 ∗ 1.0 m o l = 2.0 m o l Then, calculate the concentration of N H 3 at equilibrium: [ N H 3 ] = 5 d m 3 2.0 m o l = 0.4 M Now, calculate the equilibrium constant K c :
K c = [ N 2 ] ∗ [ H 2 ] 3 [ N H 3 ] 2 = 0.04 ∗ ( 1.8 ) 3 ( 0.4 ) 2 = 0.04 ∗ 5.832 0.16 = 0.23328 0.16 = 0.68587 Therefore, K c = 0.686
Effect of Volume Change on N H 3 Yield According to Le Chatelier's principle, if the volume of the container is decreased, the equilibrium will shift to the side with fewer moles of gas. In this reaction, there are 4 moles of gas on the reactant side ( N 2 + 3 H 2 ) and 2 moles of gas on the product side ( 2 N H 3 ). Therefore, decreasing the volume will favor the forward reaction, leading to an increase in the yield of N H 3 .
Determining if the Forward Reaction is Endothermic or Exothermic Since the equilibrium constant decreases when the temperature increases, the reverse reaction is favored at higher temperatures. This indicates that the forward reaction is exothermic.
Explanation of the Effect of Temperature on Equilibrium When the temperature of the reaction increases from 573 K to 700 K, the equilibrium constant K c decreases. This means that the equilibrium shifts towards the reactants ( N 2 and H 2 ) to counteract the increase in temperature. According to Le Chatelier's principle, if increasing the temperature favors the reverse reaction, then the forward reaction must release heat. Therefore, the forward reaction is exothermic.
Examples
The Haber-Bosch process, which synthesizes ammonia from nitrogen and hydrogen, is a crucial industrial application of this equilibrium. Understanding how temperature, pressure, and concentration affect the equilibrium allows engineers to optimize the production of ammonia, a key component in fertilizers. By manipulating these conditions, they can maximize yield and minimize energy consumption, making the process more efficient and sustainable. This optimization directly impacts global food production by ensuring a stable supply of fertilizers.
Dynamic equilibrium occurs when the forward and reverse reaction rates are equal, maintaining constant concentrations of substances. The concentrations at equilibrium are [ N 2 ] = 0.04 M and [ H 2 ] = 1.8 M , and the equilibrium constant is K c = 0.686 . The yield of N H 3 would increase if a smaller container is used, and the forward reaction is exothermic, as indicated by a decrease in K c with an increase in temperature.
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