   # Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. TiO 2 (s) + 3 C(s) → 2 CO(g) + TiC(s) (a) Calculate Δ r G ° and K at 727 °C. (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at 727 °C? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 65IL
Textbook Problem
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## Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature.TiO2(s) + 3 C(s) → 2 CO(g) + TiC(s) (a) Calculate ΔrG° and K at 727 °C. (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at 727 °C?

(a)

Interpretation Introduction

Interpretation:

The ΔrGo value and equilibrium constant for the formation of titanium carbide should be calculated.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

ΔrG°fG°(products)fG°(reactants)

ΔrGo is related to the equilibrium constant Kp by the equation,

ΔrGo=-RTlnKp

The rearranged expression is,

Kp= e-ΔrGoRT

### Explanation of Solution

The ΔrGo value and the equilibrium constant for the formation of titanium carbide are calculated below.

Given:

The Appendix L referred for the values of standard free energy values.

The given reaction is,

TiO2(s) + 3C(s)2CO(g)+TiC(s)

The ΔrGo for TiO2(s) is 757.8 kJ/mol.

The ΔrGo for C(s) is 0 kJ/mol.

The ΔrGo for CO(g) is 200.2 kJ/mol.

The ΔrGo for TiC(s) is 162.6 kJ/mol.

ΔrG°fG°(products)fG°(reactants)=[[(1 mol TiC(s)/mol-rxn)ΔfG°[TiC(s)]+(2 mol CO(g)/mol-rxn)ΔfG°[CO(g)]]-[(1 mol TiO2(s)/mol-rxn)ΔfG°[TiO2(s)]+(3 mol C(s)/mol-rxn

(b)

Interpretation Introduction

Interpretation:

It should be identified that the reaction will be product favoured at equilibrium at 727oC.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

ΔrG°fG°(products)fG°(reactants)

ΔrGo is related to the equilibrium constant Kp by the equation,

ΔrGo=-RTlnKp

The rearranged expression is,

Kp= e-ΔrGoRT

(c)

Interpretation Introduction

Interpretation:

The way how the products and reactants for given reaction adjusted in order to proceed at given temperature.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

ΔrG°fG°(products)fG°(reactants)

ΔrGo is related to the equilibrium constant Kp by the equation,

ΔrGo=-RTlnKp

The rearranged expression is,

Kp= e-ΔrGoRT

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