The equilibrium constant for the chemical equation N2(g)+3H2(g)↽−−⇀2NH3(g) N 2 ( g ) + 3 H 2 ( g ) ↽ − − ⇀ 2 NH 3 ( g ) is Kp = 0.174 0.174 at 243 243 °C. Calculate the value of Kc for the reaction at 243 243 °C.

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The equilibrium constant for the chemical equation N2(g)+3H2(g)↽−−⇀2NH3(g) N 2 ( g ) + 3 H 2 ( g ) ↽ − − ⇀ 2 NH 3 ( g ) is Kp = 0.174 0.174 at 243 243 °C. Calculate the value of Kc for the reaction at 243 243 °C.

Answer :

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Mergus Mergus · Answer 1 · 2025-03-24T11:21:26-04:00

Answer:

[tex] K_c=312.13[/tex]

Explanation:

The relation between Kp and Kc is given below:

[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

[tex]N_2_{(g)}+3H_2_{(g)}\rightleftharpoons 2NH_3_{(g)} [/tex]

Given: Kp = 0.174

Temperature = 243 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (243 + 273.15) K = 516.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

[tex]0.174= K_c\times (0.082057\times 516.15)^{-2}[/tex]

[tex] K_c\frac{1}{1793.83764}=0.174[/tex]

[tex] K_c=0.174\times \:1793.83764[/tex]

[tex] K_c=312.13[/tex]

tutorareej tutorareej · Answer 2 · 2025-03-24T11:21:17-04:00

The value of the molar equilibrium constant for the reaction has been 311.69.

The given chemical reaction has been:

[tex]\rm N_2\;+\;3\;H_2\;\rightarrow\;2\;NH_3[/tex]

The moles of reactants and products can be given by the stoichiometric coefficient.

Moles of reactant = 1 + 3 = 4

Moles of product = 2

The Pressure equilibrium constant ([tex]\rm k_p[/tex]) can be given by:

[tex]\rm k_p[/tex] = [tex]\rm k_c[/tex] [tex]\times[/tex] [tex]\rm (RT)^\Delta ^n[/tex]

[tex]\rm k_c[/tex] = Molar equilibrium constant

[tex]\rm k_p[/tex] = Pressure equilibrium constant = 0.174

R = Rydberg constant = 0.0820 L atm [tex]\rm mol^-^1\;K^-^1[/tex]

T = temperature = 243 [tex]\rm ^\circ C[/tex] = 516.15 K

[tex]\rm \Delta[/tex]n = Moles of product - moles of reactant

[tex]\rm \Delta[/tex]n = 2 - 4

[tex]\rm \Delta[/tex]n =-2

Substituting the values:

0.174 = [tex]\rm k_c[/tex] [tex]\times[/tex] [tex]\rm (0.0820\;\times\;516.15)^-^2[/tex]

0.174 = [tex]\rm k_c[/tex] [tex]\times[/tex] [tex]\rm (42.3243)^-^2[/tex]

0.174 = [tex]\rm k_c[/tex] [tex]\times[/tex] 0.00055

[tex]\rm k_c[/tex] = 311.69

The value of the molar equilibrium constant for the reaction has been 311.69.

For more information about the molar equilibrium constant, refer to the link:

https://brainly.com/question/15870179