What will be the standard internal energy change for the reaction at 298 K? OF2(g)+H2O(g) → O2(g)+2HF(g) ; Δ H° = -310 kj

Question

What will be the standard internal energy change for the reaction at 298 K? OF2(g)+H2O(g) → O2(g)+2HF(g) ; Δ H° = -310 kj (A) -312.5 kj (B) -125.03 kj (C) -310 kj (D) -156 kj

Tardigrade Admin · Accepted Answer

Δ H = Δ U + Δ ngRT Δ H = -310 × 103 J, Δ ng = 3-2=1, R= 8.314 J K-1 mol-1, T = 298 K Δ U= -310 × 10-3-(1×8.314 ×298) = -312.477 × 103 J = -312.5 kJ

Q. What will be the standard internal energy change for the reaction at $298 K$ ?
$O F_{2 (g)} + H_{2} O_{(g)} \to O_{2 (g)} + 2 H F_{(g)}; Δ H^{\circ} = - 310 kj$

$- 312.5 kj$

$- 125.03 kj$

$- 310 kj$

$- 156 kj$

$Δ H = Δ U + Δ n_{g} RT$
$Δ H = - 310 \times 1 0^{3} J$ ,
$Δ n_{g} = 3 - 2 = 1$ ,
$R = 8.314 J K^{- 1} m o l^{- 1}$ ,
$T = 298 K$
$Δ U = - 310 \times 1 0^{- 3} - (1 \times 8.314 \times 298)$
$= - 312.477 \times 1 0^{3} J = - 312.5 k J$

Q. What will be the standard internal energy change for the reaction at 298K? OF2(g)​+H2​O(g)​→O2(g)​+2HF(g)​;ΔH∘=−310kj

−312.5kj

−125.03kj

−310kj

−156kj