Calculate $\Delta$S_surr^$\omicron$at 298 K for the reaction H₂(g)+ F₂(g)$\rightarrow$ 2HF(g) $\Delta$H_r° = -546 kJ.mol^-¹, $\Delta$S_r° = +14.1 J.K^-¹.mol^-¹ A) +14.1 J.K^-¹.mol^-¹ B) +1820 J.K^-¹.mol^-¹ C) +1830 J.K^-¹.mol^-¹ D)-1830 J.K^-¹.mol^-¹ E) -14.1 J.K^-¹.mol^-¹

Question

Quizplus · Accepted Answer

The change in entropy of the surroundings ($\Delta S_{surr}^\circ$) for a reaction can be calculated using the formula $\Delta S_{surr}^\circ = -\frac{\Delta H_{r}^\circ}{T}$, where $\Delta H_{r}^\circ$ is the change in enthalpy of the reaction and $T$ is the temperature in Kelvin. Given $\Delta H_{r}^\circ = -546 \,kJ/mol$ and $T = 298 \,K$, converting $\Delta H_{r}^\circ$ to $J$ (since 1 kJ = 1000 J), we get $\Delta H_{r}^\circ = -546,000 \,J/mol$. Substituting these values into the formula gives $\Delta S_{surr}^\circ = -\frac{-546,000 \,J/mol}{298 \,K} = +1832.21 \,J/K\cdot mol$, which rounds to +1830 J/K·mol, matching choice C.

Calculate Δ\DeltaΔ Ssurr ο\omicronο at 298 K for the Reaction H2(g)+ F2(g) →\rightarrow→

Calculate $\Delta$ S_surr^$\omicron$at 298 K for the Reaction
H₂(g)+ F₂(g) $\rightarrow$