Solveeit Logo
Login

Question

Question: In which of the following \(\Delta H=\Delta U\)? (a)- \({{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(...

In which of the following ΔH=ΔU\Delta H=\Delta U?
(a)- N2(g)+3H2(g)→2NH3(g){{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)
(b)- C(s)+O2(g)→CO2(g)C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)
(c)- PCl5(g)→PCl3(g)+Cl2(g)PC{{l}_{5}}(g)\to PC{{l}_{3}}(g)+C{{l}_{2}}(g)
(d)- CaCO3(s)→CaO(s)+CO2(g)CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)

Explanation

Solution

To solve the given question we have to use the formula ΔH=ΔU+ΔnRT\Delta H=\Delta U+\Delta nRT. In the question we are not given any value like enthalpy, internal energy, and temperature, so we have to calculate the change in the number of moles in the reaction.

Complete answer:
ΔH\Delta H is the change in the enthalpy and ΔU\Delta U is the change in the internal energy of the reaction. To solve the given question we have to use the formula ΔH=ΔU+ΔnRT\Delta H=\Delta U+\Delta nRT. In the question we are not given any value like enthalpy, internal energy, and temperature, so we have to calculate the change in the number of moles in the reaction.
The change in the number of moles (Δn\Delta n) is calculated by taking the difference in the number of moles in the product and reactant.
(a)- N2(g)+3H2(g)→2NH3(g){{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)
On the product side there are 2 moles and on the reactant side there are 4 moles (3+1).
So, the change in the number of moles will be:
Δn=4−2=2\Delta n=4-2=2
(b)- C(s)+O2(g)→CO2(g)C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)
On the product side, there is 1 mole and on the reactant side, there is 1 mole.
So, the change in the number of moles will be:
Δn=1−1=0\Delta n=1-1=0
(c)- PCl5(g)→PCl3(g)+Cl2(g)PC{{l}_{5}}(g)\to PC{{l}_{3}}(g)+C{{l}_{2}}(g)
On the product side, there are 2 moles and on the reactant side, there is 1 mole.
So, the change in the number of moles will be:
Δn=2−1=1\Delta n=2-1=1
(d)- CaCO3(s)→CaO(s)+CO2(g)CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)
On the product side, there is 1 mole and on the reactant side, there is 0 mole.
So, the change in the number of moles will be:
Δn=1−0=1\Delta n=1-0=1
So, the change in the number of moles of option (b) is zero. Hence, when we put the value in the formula, we get:
ΔH=ΔU+ΔnRT\Delta H=\Delta U+\Delta nRT
ΔH=ΔU+0 x RT\Delta H=\Delta U+0\text{ x }RT
ΔH=ΔU\Delta H=\Delta U

Therefore, the correct answer is option (b)- C(s)+O2(g)→CO2(g)C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g).

Note:
When you are counting the number of moles in this type of question, only the number of moles of gaseous compounds will be considered. If the state of the compound is solid, liquid, and aqueous, then it will not be considered.