Question

6.1 a certain amount of nahco₃(s) was put into a vacuum container, then it is heated until an equilibrium was reached: 2nahco₃(s)=na₂co₃(s)+h₂o(g)+co₂(g) r=____; c=__; p=__; f=__ solution: 1 2 3 1 6.2 a certain amount of nh₄hco₃(s) was put into a vacu container, then it is heated until an equilibrium was reached: nh₄hco₃(s) ==== nh₃(g)+co₂(g)+h₂o(l) r=__; c=__; p=__; f=____。 solution: 1 2 3 1

Explanation:

6.1 Solution (Using Phase Rule Concepts)

To solve for \( R', C, P, F \) for the reaction \( \boldsymbol{2\ NaHCO_3(s) = Na_2CO_3(s) + H_2O(g) + CO_2(g)} \):

• Step 1: Determine \( R' \) (Reactions)

There is 1 independent reaction (the given decomposition), so \( R' = 1 \).

• Step 2: Determine \( C \) (Component Number)

Components are species minus independent reactions. Species: \( \text{NaHCO}_3, \text{Na}_2\text{CO}_3, \text{H}_2\text{O}, \text{CO}_2 \) (4 species). Subtract \( R' = 1 \), so \( C = 4 - 1 = 3 \)? Wait, no—alternatively, for solid - gas equilibria, solids are pure phases. Let's use the formula \( C = S - R - R' \), where \( S \) is species, \( R \) is stoichiometric reactions, \( R' \) is concentration restrictions. Here, \( \text{H}_2\text{O}(g) \) and \( \text{CO}_2(g) \) are in a 1:1 molar ratio from the reaction, so \( R' = 1 \) (concentration restriction: \( [\text{H}_2\text{O}] = [\text{CO}_2] \)). \( S = 4 \) (2 solids, 2 gases), \( R = 1 \) (the reaction). So \( C = 4 - 1 - 1 = 2 \)? Wait, the solution given is \( C = 2 \). Let's re - evaluate:
Species: \( \text{NaHCO}_3(s), \text{Na}_2\text{CO}_3(s), \text{H}_2\text{O}(g), \text{CO}_2(g) \) (4 species). The reaction is \( 2\text{NaHCO}_3 = \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \), so \( R = 1 \). The concentration of \( \text{H}_2\text{O}(g) \) and \( \text{CO}_2(g) \) are equal (from the reaction), so \( R' = 1 \) (concentration restriction). Thus, \( C = S - R - R' = 4 - 1 - 1 = 2 \).

• Step 3: Determine \( P \) (Phase Number)

Phases: solid \( \text{NaHCO}_3 \), solid \( \text{Na}_2\text{CO}_3 \), gas (mixture of \( \text{H}_2\text{O} \) and \( \text{CO}_2 \)). So \( P = 3 \).

• Step 4: Determine \( F \) (Degree of Freedom) using Phase Rule \( F = C - P + 2 \)

\( F = 2 - 3 + 2 = 1 \).

Final Answers for 6.1:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

6.2 Solution (Using Phase Rule Concepts)

For the reaction \( \boldsymbol{\text{NH}_4\text{HCO}_3(s)
ightleftharpoons \text{NH}_3(g) + \text{CO}_2(g) + \text{H}_2\text{O}(l)} \):

• Step 1: Determine \( R' \) (Reactions/Concentration Restrictions)

There is 1 independent reaction (decomposition), so \( R' = 1 \) (no additional concentration restrictions beyond stoichiometry here? Wait, the solution has \( R' = 1 \)).

• Step 2: Determine \( C \) (Component Number)

Species: \( \text{NH}_4\text{HCO}_3(s), \text{NH}_3(g), \text{CO}_2(g), \text{H}_2\text{O}(l) \) (4 species). Reaction \( R = 1 \), concentration restrictions \( R' = 1 \)? Wait, \( \text{H}_2\text{O} \) is liquid (pure phase), \( \text{NH}_3 \) and \( \text{CO}_2 \) are gases. The reaction produces \( \text{NH}_3 \) and \( \text{CO}_2 \) in 1:1 ratio, so \( R' = 1 \) (concentration restriction: \( [\text{NH}_3] = [\text{CO}_2] \)). Using \( C = S - R - R' \), \( S = 4 \), \( R = 1 \), \( R' = 1 \), so \( C = 4 - 1 - 1 = 2 \).

• Step 3: Determine \( P \) (Phase Number)

Phases: solid \( \text{NH}_4\text{HCO}_3 \), liquid \( \text{H}_2\text{O} \), gas (mixture of \( \text{NH}_3 \) and \( \text{CO}_2 \)). So \( P = 3 \).

• Step 4: Determine \( F \) (Degree of Freedom) using Phase Rule \( F = C - P + 2 \)

\( F = 2 - 3 + 2 = 1 \).

Final Answers for 6.2:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

Summary of Answers:

6.1:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

6.2:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

Answer:

For the reaction \( \boldsymbol{\text{NH}_4\text{HCO}_3(s)
ightleftharpoons \text{NH}_3(g) + \text{CO}_2(g) + \text{H}_2\text{O}(l)} \):

• Step 1: Determine \( R' \) (Reactions/Concentration Restrictions)

There is 1 independent reaction (decomposition), so \( R' = 1 \) (no additional concentration restrictions beyond stoichiometry here? Wait, the solution has \( R' = 1 \)).

• Step 2: Determine \( C \) (Component Number)

Species: \( \text{NH}_4\text{HCO}_3(s), \text{NH}_3(g), \text{CO}_2(g), \text{H}_2\text{O}(l) \) (4 species). Reaction \( R = 1 \), concentration restrictions \( R' = 1 \)? Wait, \( \text{H}_2\text{O} \) is liquid (pure phase), \( \text{NH}_3 \) and \( \text{CO}_2 \) are gases. The reaction produces \( \text{NH}_3 \) and \( \text{CO}_2 \) in 1:1 ratio, so \( R' = 1 \) (concentration restriction: \( [\text{NH}_3] = [\text{CO}_2] \)). Using \( C = S - R - R' \), \( S = 4 \), \( R = 1 \), \( R' = 1 \), so \( C = 4 - 1 - 1 = 2 \).

• Step 3: Determine \( P \) (Phase Number)

Phases: solid \( \text{NH}_4\text{HCO}_3 \), liquid \( \text{H}_2\text{O} \), gas (mixture of \( \text{NH}_3 \) and \( \text{CO}_2 \)). So \( P = 3 \).

• Step 4: Determine \( F \) (Degree of Freedom) using Phase Rule \( F = C - P + 2 \)

\( F = 2 - 3 + 2 = 1 \).

Final Answers for 6.2:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

Summary of Answers:

6.1:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)

6.2:

\( R' = \boldsymbol{1} \); \( C = \boldsymbol{2} \); \( P = \boldsymbol{3} \); \( F = \boldsymbol{1} \)