Answer:
c. Two moles of hydrogen iodide gas react to form hydrogen gas and iodine gas in a 40-L container.
Explanation:
The enthalpy (ΔH) is related to the internal energy (ΔE) through the following expressions.
ΔH = ΔE + P.ΔV [1]
ΔH = ΔE + R.T.Δn(gas) [2]
where,
P is the pressure
ΔV is the change in the volume
R is the ideal gas constant
T is the absolute temperature
Δn(gas) is the change in the gaseous moles
When ΔV = 0 or Δn(gas) = 0, ΔH will be equal to ΔE.
<em>In which of the following processes is ΔH = ΔE?
</em>
<em>a. Two moles of ammonia gas are cooled from 325 °C to 300 °C at 1.2 atm. </em>NO. When ammonia is cooled, the volume is decreased, that is, ΔV < 0 and ΔH ≠ ΔE.
<em>b. One gram of water is vaporized at 100 °C and 1 atm.</em> NO. The corresponding reaction is: H₂O(l) ⇒ H₂O(g). Δn(gas) = 1 - 0 = 1. Then, ΔH ≠ ΔE.
<em>c. Two moles of hydrogen iodide gas react to form hydrogen gas and iodine gas in a 40-L container.</em> YES. The container has a constant volume (ΔV = 0). Also, the corresponding reaction is: 2 HI(g) ⇄ H₂(g) + I₂(g). Δn(gas) = 2 - 2 = 0. Then, ΔH = ΔE.
<em>d. Calcium carbonate is heated to form calcium oxide and carbon dioxide in a container with variable volume.</em> NO. The corresponding reaction is: CaCO₃(s) ⇒ CaO(s) + CO₂(g). Δn(gas) = 1 - 0 = 1. Then, ΔH ≠ ΔE.
