Question

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) Kc = Use the following data at 751 K to find the unknown Kc: (1) N2(g) + 3 H2(g) LaTeX: \Longleftrightarrow ⟺ 2 NH3(g) Kc1 = 0.282 (2) H2(g) + I2(g) LaTeX: \Longleftrightarrow ⟺ 2 HI(g) Kc2 = 41 Enter to 0 decimal places.

Answer 1

The equilibrium constant for the given reaction at 751 K is 3.546.

Step-by-step explanation:

The equilibrium constant (Kc) for each related reaction can be found using the given information. To find the equilibrium constant for the reaction 2NH3(g) + 3I2(g) ⟶ 6HI(g) + N2(g), we can use the equilibrium constant expression from the related reactions.

First, let's find the equilibrium constant for the reaction N2(g) + 3H2(g) ⟶ 2NH3(g) (Kc1 = 0.282). The equilibrium constant expression is [NH3]^2/[N2][H2]^3. Since the given reaction is the reverse of this reaction, the equilibrium constant for the given reaction is the reciprocal of Kc1, which is 1/Kc1.

Therefore, the equilibrium constant for the reaction 2NH3(g) + 3I2(g) ⟶ 6HI(g) + N2(g) at 751 K is 1/0.282 = 3.546.

Answer 2

Answer: The equilibrium constant for the total reaction is
`4.09* 10^(-6)`

Step-by-step explanation:

We are given:

`K_(c_1)=0.282\\\\K_(c_2)=41`

We are given two intermediate equations:

Equation 1:
`N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_(c_1)=0.282`

The expression of
`K_(c_1)` for the above equation is:

`K_(c_1)=([NH_3]^2)/([N_2][H_2]^3)`

`0.282=([NH_3]^2)/([N_2][H_2]^3)` .......(1)

Equation 2:
`H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_(c_2)=41`

The expression of
`K_(c_2)` for the above equation is:

`K_(c_2)=([HI]^2)/([H_2][I_2])`

`41=([HI]^2)/([H_2][I_2])` ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

`(41)^3=([HI]^6)/([H_2]^3[I_2]^3)`

Now, dividing expression 1 by expression 2, we get:

`(K_(c_1))/(K_(c_2))=\left((([NH_3]^2)/([N_2][H_2]^3))/(([HI]^6)/([H_2]^3[l_2]^3))\right)\\\\\\(0.282)/(68921)=([NH_3]^2[I_2]^3)/([N_2][HI]^6)`

`([NH_3]^2[I_2]^3)/([N_2][HI]^6)=4.09* 10^(-6)`

The above expression is the expression for equilibrium constant of the total equation, which is:

`2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c`

Hence, the equilibrium constant for the total reaction is
`4.09* 10^(-6)`