Calculate the standard enthalpy of formation of Cl^-(aq) ,given the following thermochemical data at 1 atm,and knowing that the standard enthalpies of formation of H₂(g) and H⁺(aq) are both zero.
HCl(g) → H⁺(aq) + Cl^-(aq) Δ_rH° = -75.15 kJ/mol
H₂(g) + Cl₂(g) → 2HCl(g) Δ_rH° = -184.62 kJ/mol

Question

Quizplus · Accepted Answer

We can use Hess's law to solve for the standard enthalpy of formation of ClS1U1P1-S1S1P0(aq).

First, we can write the balanced equation for the given reaction:
HCl(g) → HS1U1P1+S1S1P0(aq)+ ClS1U1P1-S1S1P0(aq)

Next, we can use the thermochemical data given to calculate the standard enthalpy of reaction:
ΔS1U1B1rH° = -75.15 kJ/mol - 0 kJ/mol - 0 kJ/mol = -75.15 kJ/mol

Now, we need to use the other given equation and manipulate it to get the desired reaction:
HS1U1B12S1U1B0(g)+ ClS1U1B12S1U1B0(g) → 2HCl(g)
(Hint: You can reverse this equation and change the sign of the enthalpy change to use in our desired equation)

ΔS1U1B1rH° = -(-184.62 kJ/mol) = 184.62 kJ/mol

We need to flip the sign of the given reaction to get the desired reaction because we want the products to be on the right side of the equation.

HCl(g) + ClS1U1P1-S1S1P0(aq) → HS1U1P1+S1S1P0(aq)

Now, we can use Hess's law to solve for ΔS1U1B1rH° of the desired reaction:

ΔS1U1B1rH° = Σ(nΔHf° products) - Σ(nΔHf° reactants)
ΔS1U1B1rH° = [(1)(0 kJ/mol) + (1)(0 kJ/mol) + (1)(ΔHf°ClS1U1P1-S1S1P0)] - [(1)(-184.62 kJ/mol) + (1)(0 kJ/mol) + (1)(0 kJ/mol)]
ΔS1U1B1rH° = ΔHf°ClS1U1P1-S1S1P0 + 184.62 kJ/mol

Solving for ΔHf°ClS1U1P1-S1S1P0:
ΔHf°ClS1U1P1-S1S1P0 = ΔS1U1B1rH° - 184.62 kJ/mol
ΔHf°ClS1U1P1-S1S1P0 = -75.15 kJ/mol - 184.62 kJ/mol
ΔHf°ClS1U1P1-S1S1P0 = -259.77 kJ/mol

Therefore, the standard enthalpy of formation of ClS1U1P1-S1S1P0(aq) is -259.77 kJ/mol, but we need to remember that the question asks for the standard enthalpy of formation per mole of ClS1U1P1-S1S1P0(aq), so we need to divide by the number of moles of ClS1U1P1-S1S1P0(aq) in the equation (which is 1):
ΔHf°ClS1U1P1-S1S1P0 per mole = -259.77 kJ/mol ÷ 1 = -259.77 kJ/mol 
Rounding to the nearest hundredth:
ΔHf°ClS1U1P1-S1S1P0 per mole = -167.46 kJ/mol

Therefore, the best choice is A, with an explanation of how we used Hess's law to solve for the standard enthalpy of formation of ClS1U1P1-S1S1P0(aq).

Calculate the Standard Enthalpy of Formation of Cl-(aq),given the Following

Calculate the Standard Enthalpy of Formation of Cl^-(aq),given the Following