At 298 K what will be the change in standard internal energy change for the given reaction OF 2( g )+ H 2 O ( g ) longrightarrow O 2( g )+ 2 HF ( g ) Δ H =-310 kJ

Question

At 298 K what will be the change in standard internal energy change for the given reaction OF 2( g )+ H 2 O ( g ) longrightarrow O 2( g )+ 2 HF ( g ) Δ H =-310 kJ (A) - 312.5 kJ (B) - 125.03 kJ (C) - 310 kJ (D) - 156 kJ

Tardigrade Admin · Accepted Answer

Relation between internal energy change

(Δ E^{\circ})

and enthalpy change

(Δ H^{\circ})

of a reaction is given by

Δ H^{\circ} = Δ E^{\circ} + Δ n_{g} RT

...(i)

Δ n_{g} =

moles of product

-

moles of reactant For the reaction given,

O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)

Δ n_{g} = (1 + 2) - (l + l)

Δ n_{g} = l {Δ H = - 310 k J, = - 310000 J}

Putting the values in Eq (i)

- 310000 J = Δ E + l (8.3141 J K^{- 1} m o l^{- 1}) (298 K)

Δ E = - 312, 477.572 J \approx - 312.5 k J

Q. At 298K what will be the change in standard internal energy change for the given reaction OF2​(g)+H2​O(g)⟶O2​(g)+2HF(g)ΔH=−310kJ

- 312.5 kJ

- 125.03 kJ

- 310 kJ

- 156 kJ

Q. At $298 K$ what will be the change in standard internal energy change for the given reaction $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g) Δ H = - 310 k J$