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In an analysis of interhalogen reactivity, 0.440 mol of ICl was placed in a 5.00−L flask, where it decomposed at a high T: 2 ICl(g) ⇌ I2(g) + Cl2(g) Calculate the equilibrium concentrations of I2, Cl2, and ICl (Kc = 0.110 at this temperature). [I2]eq: M [Cl2]eq: M [ICl]eq: M

Respuesta :

Answer:

[I2] = [Cl2] = 0.0176 M

[ICl]  = 0.0528 M

Explanation:

Step 1: Data given

Number of moles ICI = 0.440 moles

Volume = 5.00 L

Kc = 0.110

Step 2: The balanced equation

2 ICl(g) ⇌ I2(g) + Cl2(g)

Step 3: The initial concentrations

[[ICl] = 0.440 moles/5.00 L = 0.088 M

[I2[ = 0M

[Cl2] = 0M

Step 4: The concentrations at the equilibrium

[[ICl] =(0.088 - 2x) M

[I2[ =xM

[Cl2] = xM

Step 5: Define Kc

Kc =0.110 = [I2][Cl2] / [ICl]²  

0.110 = x² / (0.088 - 2x)²

0.332 = x(0.088 - 2x)

x = 0.332 *(0.088 - 2x)

x = 0.029216 - 0.664x

1.664x = 0.029216

x = 0.0176

[I2] = [Cl2] = 0.0176 M

[ICl] = 0.088  - 2(0.0176 )  = 0.0528 M

The equilibrium concentrations of

[I₂] = [Cl₂] = 0.0176 M

[ICl]  = 0.0528 M

Given:

Number of moles ICI = 0.440 moles

Volume = 5.00 L

Kc = 0.110

Balanced chemical equation:

2 ICl(g) ⇌ I₂(g) + Cl₂(g)

Calculation for initial concentration:

[ICl] = 0.440 moles/5.00 L = 0.088 M

[I₂] = 0M

[Cl₂] = 0M

Concentrations at equilibrium:

[ICl] =(0.088 - 2x) M

[I₂] =xM

[Cl₂] = xM

Calculation for Kc:

Kc =0.110 = [I₂][Cl₂] / [ICl]²  

0.110 = x² / (0.088 - 2x)²

0.332 = x(0.088 - 2x)

x = 0.332 *(0.088 - 2x)

x = 0.029216 - 0.664x

1.664x = 0.029216

x = 0.0176

Thus, the equilibrium concentrations  will be:

[I₂] = [Cl₂] = 0.0176 M

[ICl] = 0.088  - 2(0.0176 )  = 0.0528 M

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