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To calculate the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\( \text{HCl} (g) + \text{NaNO}_2(s) \rightarrow \text{HNO}_2 (l) + \text{NaCl} (s) \)[/tex] using Hess's Law, we need to manipulate the given reactions to match the target reaction.
Given data:
1. [tex]\( 2 \text{NaCl} (s) + \text{H}_2\text{O} (l) \rightarrow 2 \text{HCl} (g) + \text{Na}_2\text{O} (s) \)[/tex] [tex]\[\Delta H_1 = 382.0 \text{ kJ}\][/tex]
2. [tex]\( \text{NO} (g) + \text{NO}_2(g) + \text{Na}_2\text{O} (s) \rightarrow 2 \text{NaNO}_2(s) \)[/tex] [tex]\[\Delta H_2 = -601.0 \text{ kJ}\][/tex]
3. [tex]\( \text{NO} (g) + \text{NO}_2(g) \rightarrow \text{N}_2\text{O} (g) + \text{O}_2(g) \)[/tex] [tex]\[\Delta H_3 = -40.0 \text{ kJ}\][/tex]
4. [tex]\( 2 \text{HNO}_2(l) \rightarrow \text{N}_2\text{O} (g) + \text{O}_2(g) + \text{H}_2\text{O} (l) \)[/tex] [tex]\[\Delta H_4 = 29.0 \text{ kJ}\][/tex]
We need the target reaction:
[tex]\[ \text{HCl} (g) + \text{NaNO}_2(s) \rightarrow \text{HNO}_2 (l) + \text{NaCl} (s) \][/tex]
Let's manipulate the given reactions step-by-step to achieve the target reaction.
### Step 1: Reverse reaction 1 and divide by 2
[tex]\[ 2 \text{HCl} (g) + \text{Na}_2\text{O} (s) \rightarrow 2 \text{NaCl} (s) + \text{H}_2\text{O} (l) \][/tex]
[tex]\[ \Delta H_1' = -382.0 \text{ kJ}\][/tex]
Dividing by 2:
[tex]\[ \text{HCl} (g) + \frac{1}{2} \text{Na}_2\text{O} (s) \rightarrow \text{NaCl} (s) + \frac{1}{2} \text{H}_2\text{O} (l) \][/tex]
[tex]\[ \Delta H_1'' = \frac{-382.0}{2} = -191.0 \text{ kJ}\][/tex]
### Step 2: Use reaction 2 as is
[tex]\[ \text{NO} (g) + \text{NO}_2(g) + \frac{1}{2} \text{Na}_2\text{O} (s) \rightarrow \text{NaNO}_2(s) \) \[ \Delta H_2 = -601.0 \text{ kJ}\][/tex]
### Step 3: Reverse reaction 3
[tex]\[ \text{N}_2\text{O} (g) + \text{O}_2(g) \rightarrow \text{NO} (g) + \text{NO}_2(g) \) \[ \Delta H_3' = 40.0 \text{ kJ}\][/tex]
### Step 4: Reverse reaction 4
[tex]\[ \text{N}_2\text{O} (g) + \text{O}_2(g) + \text{H}_2\text{O} (l) \rightarrow 2 \text{HNO}_2(l) \) \[ \Delta H_4' = -29.0 \text{ kJ}\][/tex]
### Combine These Steps
1. [tex]\(\Delta H_1'' = -191.0 \text{ kJ}\)[/tex]
2. [tex]\(\Delta H_2 = -601.0 \text{ kJ}\)[/tex]
3. [tex]\(\Delta H_3' = 40.0 \text{ kJ}\)[/tex]
4. [tex]\(\Delta H_4' = -29.0 \text{ kJ}\)[/tex]
Adding them together to get the enthalpy change for the target reaction:
[tex]\[ \Delta H_{\text{target}} = (-191.0) + (-601.0) + 40.0 + (-29.0) \][/tex]
[tex]\[ \Delta H_{\text{target}} = -781.0 \text{ kJ}\][/tex]
Thus, the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\( \text{HCl} (g) + \text{NaNO}_2(s) \rightarrow \text{HNO}_2 (l) + \text{NaCl} (s) \)[/tex] is [tex]\(-781.0 \text{ kJ}\)[/tex].
Given data:
1. [tex]\( 2 \text{NaCl} (s) + \text{H}_2\text{O} (l) \rightarrow 2 \text{HCl} (g) + \text{Na}_2\text{O} (s) \)[/tex] [tex]\[\Delta H_1 = 382.0 \text{ kJ}\][/tex]
2. [tex]\( \text{NO} (g) + \text{NO}_2(g) + \text{Na}_2\text{O} (s) \rightarrow 2 \text{NaNO}_2(s) \)[/tex] [tex]\[\Delta H_2 = -601.0 \text{ kJ}\][/tex]
3. [tex]\( \text{NO} (g) + \text{NO}_2(g) \rightarrow \text{N}_2\text{O} (g) + \text{O}_2(g) \)[/tex] [tex]\[\Delta H_3 = -40.0 \text{ kJ}\][/tex]
4. [tex]\( 2 \text{HNO}_2(l) \rightarrow \text{N}_2\text{O} (g) + \text{O}_2(g) + \text{H}_2\text{O} (l) \)[/tex] [tex]\[\Delta H_4 = 29.0 \text{ kJ}\][/tex]
We need the target reaction:
[tex]\[ \text{HCl} (g) + \text{NaNO}_2(s) \rightarrow \text{HNO}_2 (l) + \text{NaCl} (s) \][/tex]
Let's manipulate the given reactions step-by-step to achieve the target reaction.
### Step 1: Reverse reaction 1 and divide by 2
[tex]\[ 2 \text{HCl} (g) + \text{Na}_2\text{O} (s) \rightarrow 2 \text{NaCl} (s) + \text{H}_2\text{O} (l) \][/tex]
[tex]\[ \Delta H_1' = -382.0 \text{ kJ}\][/tex]
Dividing by 2:
[tex]\[ \text{HCl} (g) + \frac{1}{2} \text{Na}_2\text{O} (s) \rightarrow \text{NaCl} (s) + \frac{1}{2} \text{H}_2\text{O} (l) \][/tex]
[tex]\[ \Delta H_1'' = \frac{-382.0}{2} = -191.0 \text{ kJ}\][/tex]
### Step 2: Use reaction 2 as is
[tex]\[ \text{NO} (g) + \text{NO}_2(g) + \frac{1}{2} \text{Na}_2\text{O} (s) \rightarrow \text{NaNO}_2(s) \) \[ \Delta H_2 = -601.0 \text{ kJ}\][/tex]
### Step 3: Reverse reaction 3
[tex]\[ \text{N}_2\text{O} (g) + \text{O}_2(g) \rightarrow \text{NO} (g) + \text{NO}_2(g) \) \[ \Delta H_3' = 40.0 \text{ kJ}\][/tex]
### Step 4: Reverse reaction 4
[tex]\[ \text{N}_2\text{O} (g) + \text{O}_2(g) + \text{H}_2\text{O} (l) \rightarrow 2 \text{HNO}_2(l) \) \[ \Delta H_4' = -29.0 \text{ kJ}\][/tex]
### Combine These Steps
1. [tex]\(\Delta H_1'' = -191.0 \text{ kJ}\)[/tex]
2. [tex]\(\Delta H_2 = -601.0 \text{ kJ}\)[/tex]
3. [tex]\(\Delta H_3' = 40.0 \text{ kJ}\)[/tex]
4. [tex]\(\Delta H_4' = -29.0 \text{ kJ}\)[/tex]
Adding them together to get the enthalpy change for the target reaction:
[tex]\[ \Delta H_{\text{target}} = (-191.0) + (-601.0) + 40.0 + (-29.0) \][/tex]
[tex]\[ \Delta H_{\text{target}} = -781.0 \text{ kJ}\][/tex]
Thus, the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\( \text{HCl} (g) + \text{NaNO}_2(s) \rightarrow \text{HNO}_2 (l) + \text{NaCl} (s) \)[/tex] is [tex]\(-781.0 \text{ kJ}\)[/tex].
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