Chapter 2: Q12.6 40P (page 36)

Using the data in Appendix D, calculate $∆ H^{\circ}$ for each of the following processes:
(a) $2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$
(b) $C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$
(c) $2 {NH}_{3} (g) + \frac{O}{2} (g) \to 2 {NO}_{2} (g) + 3 H_{2} O (g)$
(d) $C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

Short Answer

Expert verified

(a) The standard enthalpy change $∆ H_{a}^{\circ} = - 114.14 k J$ .

(b) The standard enthalpy change $∆ H_{b}^{\circ} = + 172.47 k J$ .

(d) The standard enthalpy change $∆ H_{d}^{\circ} = + 131.30 k J$ .

Step by step solution

Given data

The given chemical equations are:

(a) $2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$

(b) $C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

(d) $C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

$∆ H_{f}^{\circ}$ Value of ${NO}_{2} = 33.18 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $NO = 90.25 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $O_{2} = 0$ .

$∆ H_{f}^{\circ}$ Value of $CO = - 110.52 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $C (s) = 0$ .

$∆ H_{f}^{\circ}$ Value of CO₂=-393.51kJmol^-1.

$∆ H_{f}^{\circ}$ Represents the standard enthalpy of molecules.

Step 2: Concept of Enthalpy

Enthalpy change is amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in chemical reaction.

It is represented by $∆ H$ .

Negative enthalpy change represents the exothermic reaction when energy is released from the reaction and positive enthalpy change represents an endothermic reaction in which energy is taken in from surroundings.

Formula to calculate ∆H∘

The enthalpy change of reaction is as:

$∆ H^{\circ} = n ∆ H_{f}^{\circ} (P r o d u c t s) - n ∆ H_{f}^{\circ} (R e a c \tan t s)$ ...... (1)

Where, $∆ H_{f}^{\circ}$ represents the standard enthalpy of molecules.

Calculate the value of ∆H∘

(a)

The given expression is:

$2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$

$∆ H_{f}^{\circ}$ Value of ${NO}_{2} = 33.18 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $NO = 90.25 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $O_{2} = 0$ .

$∆ H_{f}^{\circ}$ Represents the standard enthalpy of molecules.

Substitute these values in equation (1):

localid="1663776205823" $∆ {H_{a}}^{\circ} = (2 m o l) ∆ H_{f}^{\circ} [N O_{2} (g)] - (2 m o l) ∆ H_{f}^{\circ} [N O (g)] + (1 m o l) ∆ H_{f}^{\circ} [O_{2}] = (2 m o l) (33.18 k J m o l^{- 1}) - (2 m o l) (90.25 k J m o l^{- 1}) + 0 = 66.36 k J - 180.50 k J = - 114.14 k J$

Calculate the value of ∆H∘

(b)

The given chemical reaction is $C (s) + {CO}_{2} (g) \to 2 CO (g)$ .

$∆ H_{f}^{\circ}$ Value of $CO = - 110.52 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $C (s) = 0$ .

$∆ H_{f}^{\circ}$ Value of ${CO}_{2} = - 393.51 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Represents the standard enthalpy of molecules.

Substitute these values in equation (1):

$∆ {H_{b}}^{\circ} = (2 m o l) ∆ H_{f}^{\circ} [C O (g)] - (1 m o l) ∆ H_{f}^{\circ} [C (s)] + (1 m o l) ∆ H_{f}^{\circ} [C O_{2}] = (2 m o l) (- 110.52 k J m o l^{- 1}) - (0 + (1 m o l) (- 393.51 k J m o l^{- 1})) = - 221.04 k J + 393.51 k J = + 172.47 k J$

Calculate the value of ΔH∘

(c)

The given expression is:

$2 {NH}_{3} (g) + \frac{7}{2} O_{2} (g) \to 2 {NO}_{2} (g) + 3 H_{2} O (l)$

$∆ H_{f}^{\circ}$ Value of ${NO}_{2} = 33.18 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $H_{2} O = - 285.83 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Value of $O_{2} = 0$ .

$∆ H_{f}^{\circ}$ Value of ${NH}_{3} = - 46.11 {kJmol}^{- 1}$ .

$∆ H_{f}^{\circ}$ Represents the standard enthalpy of molecules.

Substitute these values in equation (1):

$∆ {H_{c}}^{\circ} = (2 m o l) ∆ H_{f}^{\circ} [N O_{2} (g)] + (3 m o l) ∆ H_{f}^{\circ} [H_{2} O (l)] - {(2 m o l) ∆ H_{f}^{\circ} [N H_{3} (g)] + (7 / 2 m o l) ∆ H_{f}^{\circ} [O_{2}] = (2 m o l) (33.18 k J m o l^{- 1}) = (3 m o l) (- 285.83 k J m o l^{- 1}) - (2 m o l) (- 46.11 k J m o l^{- 1}) \times 0 = 66.36 k J - 725.46 k J + 92.2 k J = - 566.88 k J$

Calculate the value of ∆H∘

(d)

The given expression is:

$C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

$∆ H_{f}^{\circ}$ Value of H₂O = -285.83kJmol^-1.

$∆ H_{f}^{\circ}$ Value of CO=-110.52kJmol^-1.

$∆ H_{f}^{\circ}$ Value of $H_{2} = 0$ .

$∆ H_{f}^{\circ}$ Value of C(s) =0

$∆ H_{f}^{\circ}$ Represents the standard enthalpy of molecules.

Substitute these values in equation (1):

$∆ {H_{d}}^{\circ} = (1 m o l) ∆ H_{f}^{\circ} [C O_{} (g)] + (1 m o l) ∆ H_{f}^{\circ} [H_{2} (g)] - {(1 m o l) ∆ H_{f}^{\circ} [C (s)] + (1 m o l) ∆ H_{f}^{\circ} [H_{2} O] = (1 m o l) (- 110.52 k J m o l^{- 1}) + (1 m o l) \times 0 - (1 m o l) (- 285.83 k J m o l^{- 1}) + (1 m o l) \times 0 = - 110.52 k J + 241.82 k J = + 131.30 k J$

Short Answer

Step by step solution

Given data

Step 2: Concept of Enthalpy

Formula to calculate ∆H∘

Calculate the value of ∆H∘

Calculate the value of ∆H∘

Calculate the value of ΔH∘

Calculate the value of ∆H∘

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Kinetics

Making Measurements

Physical Chemistry

Chemistry Branches

Ionic and Molecular Compounds

Study anywhere. Anytime. Across all devices.

Company

Product

Help

Using the data in Appendix D, calculate ∆H∘for each of the following processes:(a)2NO(g)+O2(g)→2NO2(g)(b)C(s)+H2O(g)→CO(g)+H2(g)(c)2NH3(g)+O2(g)→2NO2(g)+3H2O(g)(d)C(s)+H2O(g)→CO(g)+H2(g)

Short Answer

Step by step solution

Given data

Step 2: Concept of Enthalpy

Formula to calculate ∆H∘

Calculate the value of ∆H∘

Calculate the value of ∆H∘

Calculate the value of ΔH∘

Calculate the value of ∆H∘

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Kinetics

Making Measurements

Physical Chemistry

Chemistry Branches

Ionic and Molecular Compounds

Study anywhere. Anytime. Across all devices.