QuestionJuly 6, 2025

Determine the enthalpy of reaction for HCl(g) +NaNO_(2)(s)arrow HNO_(2)(l)+NaCl(s) 2NaCl(s)+H_(2)O(l)arrow 2HCl(g)+Na_(2)O(s)Delta H^circ = -507.1kJ/mol NO(g)+NO_(2)(g)+Na_(2)O(s)arrow 2NaNO_(2)(s)Delta H^circ = -427.0kJ/mol NO(g)+NO_(2)(g)arrow N_(2)O(g)+O_(2)(g)Delta H^circ = -43.01kJ/mol 2HNO_(2)(l)arrow N_(2)O(g)+O_(2)(g)+H_(2)O(l)Delta H^circ = +34.02kJ/mol

Determine the enthalpy of reaction for HCl(g) +NaNO_(2)(s)arrow HNO_(2)(l)+NaCl(s) 2NaCl(s)+H_(2)O(l)arrow 2HCl(g)+Na_(2)O(s)Delta H^circ = -507.1kJ/mol NO(g)+NO_(2)(g)+Na_(2)O(s)arrow 2NaNO_(2)(s)Delta H^circ = -427.0kJ/mol NO(g)+NO_(2)(g)arrow N_(2)O(g)+O_(2)(g)Delta H^circ = -43.01kJ/mol 2HNO_(2)(l)arrow N_(2)O(g)+O_(2)(g)+H_(2)O(l)Delta H^circ = +34.02kJ/mol
Determine the enthalpy of reaction for HCl(g)
+NaNO_(2)(s)arrow HNO_(2)(l)+NaCl(s)
2NaCl(s)+H_(2)O(l)arrow 2HCl(g)+Na_(2)O(s)Delta H^circ =
-507.1kJ/mol
NO(g)+NO_(2)(g)+Na_(2)O(s)arrow 2NaNO_(2)(s)Delta H^circ =
-427.0kJ/mol
NO(g)+NO_(2)(g)arrow N_(2)O(g)+O_(2)(g)Delta H^circ =
-43.01kJ/mol
2HNO_(2)(l)arrow N_(2)O(g)+O_(2)(g)+H_(2)O(l)Delta H^circ =
+34.02kJ/mol

Solution
4.4(82 votes)

Answer

-3.07 \text{ kJ/mol} Explanation 1. Write the target reaction The target reaction is HCl(g) + NaNO_2(s) \rightarrow HNO_2(l) + NaCl(s). 2. Reverse and adjust given reactions Reverse the second reaction to form 2NaNO_2(s) from Na_2O(s), which changes \Delta H^{\circ} to +427.0 kJ/mol. Reverse the third reaction to form NO(g) + NO_2(g) from N_2O(g) + O_2(g), changing \Delta H^{\circ} to +43.01 kJ/mol. 3. Combine reactions Add the adjusted reactions: - 2NaCl(s) + H_2O(l) \rightarrow 2HCl(g) + Na_2O(s) with \Delta H^{\circ} = -507.1 kJ/mol - 2NaNO_2(s) \rightarrow NO(g) + NO_2(g) + Na_2O(s) with \Delta H^{\circ} = +427.0 kJ/mol - N_2O(g) + O_2(g) \rightarrow NO(g) + NO_2(g) with \Delta H^{\circ} = +43.01 kJ/mol - 2HNO_2(l) \rightarrow N_2O(g) + O_2(g) + H_2O(l) with \Delta H^{\circ} = +34.02 kJ/mol 4. Simplify and calculate total enthalpy change Cancel out common species and simplify to get the target reaction. Calculate the total \Delta H^{\circ} by summing the enthalpies: \Delta H^{\circ}_{\text{total}} = -507.1 + 427.0 + 43.01 + 34.02 = -3.07 \text{ kJ/mol}

Explanation

1. Write the target reaction<br /> The target reaction is $HCl(g) + NaNO_2(s) \rightarrow HNO_2(l) + NaCl(s)$.<br /><br />2. Reverse and adjust given reactions<br /> Reverse the second reaction to form $2NaNO_2(s)$ from $Na_2O(s)$, which changes $\Delta H^{\circ}$ to $+427.0$ kJ/mol.<br /> Reverse the third reaction to form $NO(g) + NO_2(g)$ from $N_2O(g) + O_2(g)$, changing $\Delta H^{\circ}$ to $+43.01$ kJ/mol.<br /><br />3. Combine reactions<br /> Add the adjusted reactions:<br />- $2NaCl(s) + H_2O(l) \rightarrow 2HCl(g) + Na_2O(s)$ with $\Delta H^{\circ} = -507.1$ kJ/mol<br />- $2NaNO_2(s) \rightarrow NO(g) + NO_2(g) + Na_2O(s)$ with $\Delta H^{\circ} = +427.0$ kJ/mol<br />- $N_2O(g) + O_2(g) \rightarrow NO(g) + NO_2(g)$ with $\Delta H^{\circ} = +43.01$ kJ/mol<br />- $2HNO_2(l) \rightarrow N_2O(g) + O_2(g) + H_2O(l)$ with $\Delta H^{\circ} = +34.02$ kJ/mol<br /><br />4. Simplify and calculate total enthalpy change<br /> Cancel out common species and simplify to get the target reaction. Calculate the total $\Delta H^{\circ}$ by summing the enthalpies:<br />$$\Delta H^{\circ}_{\text{total}} = -507.1 + 427.0 + 43.01 + 34.02 = -3.07 \text{ kJ/mol}$$
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