Question

Show how the equations listed below with dHo values can be manipulated to calculate the dHo of the following equation

2KCl(s) + H2SO4 (l) → 2HCl (g) + K2SO4 (s) dHo = ?
a.HCl (g) + KOH (s) → KCl (s) + H2O (l) dHo = - 200.0 kJ
b.H2SO4 (l) + 2KOH (s) → K2SO4 (s) + 2 H2O (l) dHo = -340.0 kJ

Answer 1

a. HCl (g) + KOH (s) → KCl (s) + H2O (l) with ΔH° = -200.0 kJ

b. H2SO4 (l) + 2KOH (s) → K2SO4 (s) + 2 H2O (l) with ΔH° = -340.0 kJ

The sum of the enthalpy changes for the manipulated equations is ΔH° = -540.0 kJ.

Step-by-step explanation:

The given equations are:

a. HCl (g) + KOH (s) → KCl (s) + H2O (l) with ΔH° = -200.0 kJ

b. H2SO4 (l) + 2KOH (s) → K2SO4 (s) + 2 H2O (l) with ΔH° = -340.0 kJ

To find the enthalpy change (ΔH°) for the target equation 2KCl (s) + H2SO4 (l) → 2HCl (g) + K2SO4 (s), we can manipulate and combine the given equations:

Multiply the first equation by 2:

2(HCl (g) + KOH (s) → KCl (s) + H2O (l))

2HCl (g) + 2KOH (s) → 2KCl (s) + 2H2O (l)

Now add the second equation:

2HCl (g) + 2KOH (s) + H2SO4 (l) + 2KOH (s) → 2KCl (s) + 2H2O (l) + K2SO4 (s)

Combine like terms:

2HCl (g) + H2SO4 (l) + 4KOH (s) → 2KCl (s) + 3H2O (l) + K2SO4 (s)

Now, compare this to the target equation:

2KCl (s) + H2SO4 (l) → 2HCl (g) + K2SO4 (s)

The enthalpy change (ΔH°) for the target equation is the sum of the enthalpy changes of the manipulated equations:

ΔH° = (-200.0 kJ) + (-340.0 kJ) = -540.0 kJ

Therefore, the enthalpy change for the target reaction is ΔH° =-540.0 kJ.