Question

Calculate the difference between change in enthalpy and
change in internal energy of the system at 300 K for the
following reaction
C(s) + 2H2(g) → CH4(9) (R = 8.314 J/K/mol)​

Answer 1

Answer:

Solution:-

OF2(g)+H2O(g)⟶O2(g)+2HF(g)

ΔHR=∑ΔHf(product)−∑ΔHf(reactant)

∴ΔHR=(2×(−268.6))−(23+(−241.8))

⇒ΔHR=−756kJ/mol

Hence the standard enthalpy change will be −756kJ/mol

Now, as we know that,

ΔH=ΔU+ΔngRT

⇒ΔU=ΔH−ΔngRT

Now from the given reaction,

Δng=nP−nR=(2+1)−(1+1)=1

T=300K(Given)