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Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Respuesta :

Answer: The value of [tex]K_c[/tex] for the reaction is [tex]6.005\times 10^{-6}[/tex]

Explanation:

We are given:

Initial moles of [tex]NH_3=0.0150mol[/tex]

Initial moles of [tex]O_2=0.0150mol[/tex]

Volume of the container = 1.00 L

Molarity of the solution = [tex]\frac{\text{Number of moles}}{\text{Volume of container}}[/tex]

[tex][NH_3]_i=\frac{0.0150}{1.00}=0.0150M[/tex]

[tex][O_2]_i=\frac{0.0150}{1.00}=0.0150M[/tex]

The given chemical equation follows:

                     [tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)[/tex]

Initial:          0.0150        0.0150

At eqllm:   0.0150-4x     0.0150-3x    2x       6x

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}[/tex]         .......(1)

We are given:

Equilibrium concentration of [tex]N_2=1.96\times 10^{-3}[/tex]

Equating the equilibrium concentrations of nitrogen, we get:

[tex]2x=1.96\times 10^{-3}\\\\x=0.98\times 10^{-3}M[/tex]

Calculating the equilibrium concentrations:

Concentration of [tex]NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M[/tex]

Concentration of [tex]O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M[/tex]

Concentration of [tex]N_2=2x=2(0.00098)=0.00196M[/tex]

Concentration of [tex]H_2O=6x=6(0.00098)=0.00588M[/tex]

Putting values in expression 1, we get:

[tex]K_c=\frac{(0.00196)^2\times (0.00588)^6}{(0.01108)^4\times (0.01206)^3}\\\\K_c=6.005\times 10^{-6}[/tex]

Hence, the value of [tex]K_c[/tex] for the reaction is [tex]6.005\times 10^{-6}[/tex]

mahima6824soni

Kc is the ratio of equilibrium constant of product to the reactants. The value of Kc for the reaction is [tex]\rm Kc =6.005\times10^-^6[/tex]

What is Kc?

Kc is the ratio of the equilibrium constant of the product to the equilibrium constant of the reactants.

Now, The reaction is

[tex]\rm 4 NH_3(g) +3 O_2 (g) <--->2 N_2(g) + 6 H_2O(g)[/tex]

The given data is

Initial moles of  [tex]\rm NH_3[/tex] is 0.0150 mol

Initial moles of [tex]\rm O_2[/tex] is  0.0150 mol

Volume of container is 1.00−L

Step1: find the molarity of the solution

[tex]\rm M = \dfrac{n}{V}\\\\\rm NH_3 = \dfrac{0.0150 mol }{1.00L }=0.0150 \;M\\\\\\\rm O_2 = \dfrac{0.0150 mol }{1.00L }=0.0150 \;M[/tex]

Step2: Making the ICE table:

4 moles of ammonia and 3 moles of oxygen is used to prepare 2 moles of nitrogen and 6 moles of water.

Thus, we are reducing the moles from the reactants

                              [tex]\rm 4 NH_3(g) +3 O_2 (g) <--->2 N_2(g) + 6 H_2O(g)[/tex]

          Initial:          0.0150        0.0150

At equilibrium:   0.0150-4x     0.0150-3x    2x       6x

Equilibrium constant of [tex]\rm N_2\;\;is\;\; 1.96 \times 10^-^3 M[/tex]

So, [tex]\rm N_2[/tex] is 2x

[tex]\rm 2x = 1.96 \times 10^-^3\;M\\\\\rm x = 0.98\times 10^-^3\;M\\[/tex]

Step3: Calculating the equilibrium concentration

Concentration of [tex]\rm NH_3[/tex]   = [tex]0.0150-4x = 0.0150 - 4(0.00098) = 0.01108\;M[/tex]

Concentration of [tex]\rm O_2[/tex]  = [tex]0.0150-3x = 0.0150 - 3(0.00098) = 0.01206\;M[/tex]

Concentration of [tex]N_2 = 2x = 2(0.00098) = 0.00196\;M[/tex]

Concentration of [tex]H_2O = 6x= 6(0.00098) = 0.00588\;M[/tex]

Step4: Calculating the Kc

[tex]\rm Kc = \dfrac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}[/tex]

[tex]\rm Kc = \dfrac{[0.00196]^2\;[0.00588]^6}{[0.01108]^4\;[0.01206]^3} =6.005\times10^-^6[/tex]

Thus, the value of [tex]\rm Kc =6.005\times10^-^6[/tex]

Learn more about Kc, here:

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