Question

Calculate h in kj for the following reaction: HCl + NaNO₂ > HNO₂ + NaCl. Given the following equations and their enthalpy values: 3N₂O + 3O₂ > 3NO₂, ΔH = -128.04 kj; H₂O + 2 NaCl > 2HCl + Na₂O, ΔH = 507.31 kj; 1/2 no + 1/2 NO₂ + 1/2 Na₂O > NaNO₂, ΔH = -213.57 kj; 2HNO₂ > N₂O + O₂ + H₂O, ΔH = 34.35 kj. Use these individual equations and show how they are manipulated to end with the original equation?

Answer 1

The enthalpy change (ΔH) for the target reaction HCl + NaNO₂ → HNO₂ + NaCl is calculated by manipulating and summing the enthalpy changes of the given reactions to match the target equation.

Step-by-step explanation:

To calculate the change in enthalpy (ΔH) for the reaction HCl + NaNO₂ → HNO₂ + NaCl, we must manipulate and sum the enthalpy changes (ΔH) of the given reactions so that, when added, they result in the enthalpy change of the target reaction.

1. First, we reverse the second reaction and divide it by two to match the coefficients in the target equation, which requires enthalpy to change the sign and be halved:
   
2. Next, we reverse the fourth reaction and multiply it by ½ to match the coefficients in the target equation:
   
3. Regarding the third equation, it already produces NaNO₂ as required in the target equation, so no changes are needed:
   
4. Finally, divide the first equation by three to match the coefficients in the N₂O and O₂ products:
   

Summing the ΔHs of the four manipulated reactions, we get the total ΔH:

ΔH = (-507.31/2) - (34.35/2) - 213.57 - (-128.04/3) kJ