Question

determine the enthalpy of the reaction of
$ce{hcl(g) + nano_{2}(s) -> hno_{2}(l) + nacl(s)}$
using the equations:
$ce{2nacl(s) + h_{2}o(l) -> 2hcl(g) + na_{2}o(s)}$ $delta h^circ = -507.1 \text{kj/mol}$
$ce{no(g) + no_{2}(g) + na_{2}o(s) -> 2nano_{2}(s)}$ $delta h^circ = -427.0 \text{kj/mol}$
$ce{no(g) + no_{2}(g) -> n_{2}o(g) + o_{2}(g)}$ $delta h^circ = -43.01 \text{kj/mol}$
$ce{2hno_{2}(l) -> n_{2}o(g) + o_{2}(g) + h_{2}o(l)}$ $delta h^circ = +34.02 \text{kj/mol}$

Explanation:

Step1: Label given reactions

Let the target reaction be:

$$\text{HCl}(g) + \text{NaNO}_2(s) ightarrow \text{HNO}_2(l) + \text{NaCl}(s) \quad \Delta H^\circ = ?$$

Label the provided reactions:

1. $2\text{NaCl}(s) + \text{H}_2\text{O}(l)

ightarrow 2\text{HCl}(g) + \text{Na}_2\text{O}(s) \quad \Delta H_1^\circ = -507.1\ \text{kJ/mol}$

1. $\text{NO}(g) + \text{NO}_2(g) + \text{Na}_2\text{O}(s)

ightarrow 2\text{NaNO}_2(s) \quad \Delta H_2^\circ = -427.0\ \text{kJ/mol}$

1. $\text{NO}(g) + \text{NO}_2(g)

ightarrow \text{N}_2\text{O}(g) + \text{O}_2(g) \quad \Delta H_3^\circ = -43.01\ \text{kJ/mol}$

1. $2\text{HNO}_2(l)

ightarrow \text{N}_2\text{O}(g) + \text{O}_2(g) + \text{H}_2\text{O}(l) \quad \Delta H_4^\circ = +34.02\ \text{kJ/mol}$

Step2: Reverse and scale reaction 1

Reverse reaction 1 and divide by 2 to get $\text{HCl}(g)$ and $\text{NaCl}(s)$ in correct positions/amounts:

$$\text{HCl}(g) + \frac{1}{2}\text{Na}_2\text{O}(s) ightarrow \text{NaCl}(s) + \frac{1}{2}\text{H}_2\text{O}(l) \quad \Delta H_1'^\circ = \frac{+507.1}{2} = 253.55\ \text{kJ/mol}$$

Step3: Reverse and scale reaction 2

Reverse reaction 2 and divide by 2 to get $\text{NaNO}_2(s)$ as a reactant:

$$\text{NaNO}_2(s) ightarrow \frac{1}{2}\text{NO}(g) + \frac{1}{2}\text{NO}_2(g) + \frac{1}{2}\text{Na}_2\text{O}(s) \quad \Delta H_2'^\circ = \frac{+427.0}{2} = 213.5\ \text{kJ/mol}$$

Step4: Reverse reaction 3

Reverse reaction 3 to cancel $\text{NO}/\text{NO}_2$:

$$\text{N}_2\text{O}(g) + \text{O}_2(g) ightarrow \text{NO}(g) + \text{NO}_2(g) \quad \Delta H_3'^\circ = +43.01\ \text{kJ/mol}$$

Step5: Reverse and scale reaction 4

Reverse reaction 4 and divide by 2 to get $\text{HNO}_2(l)$ as a product:

$$\frac{1}{2}\text{N}_2\text{O}(g) + \frac{1}{2}\text{O}_2(g) + \frac{1}{2}\text{H}_2\text{O}(l) ightarrow \text{HNO}_2(l) \quad \Delta H_4'^\circ = \frac{-34.02}{2} = -17.01\ \text{kJ/mol}$$

Step6: Sum all modified reactions

Add the modified reactions and their enthalpies:
$$\Delta H^\circ = \Delta H_1'^\circ + \Delta H_2'^\circ + \frac{1}{2}\Delta H_3'^\circ + \Delta H_4'^\circ$$
$$\Delta H^\circ = 253.55 + 213.5 + \frac{43.01}{2} - 17.01$$

Step7: Calculate final enthalpy

Compute the total value:
$$\Delta H^\circ = 253.55 + 213.5 + 21.505 - 17.01 = 471.545\ \text{kJ/mol}$$

Answer:

$\Delta H^\circ = +471.5\ \text{kJ/mol}$ (rounded to one decimal place, or $+471.55\ \text{kJ/mol}$ for precise value)