The enthalpy and entropy change for the reaction: Br2(ℓ) + Cl2(g) arrow 2BrCl(g) are 30 kJmol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is :

Question

The enthalpy and entropy change for the reaction: Br2(ℓ) + Cl2(g) arrow 2BrCl(g) are 30 kJmol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is : (A) 285.7 K (B) 373 K (C) 250 K (D) 400 K

Tardigrade Admin · Accepted Answer

For the reaction Br 2(ℓ)+ Cl 2( g ) arrow 2 BrCl ( g ) Δ H =30 kJ / mol Δ S =105 JK -1 mol -1 For at equilibrium Δ G =0 ∴ Δ G =Δ H - T Δ S Δ H = T Δ S T =(Δ H /Δ S )=(30 × 1000 J mol -1/105 JK -1 mol -1) =2857 K

Q. The enthalpy and entropy change for the reaction : Br2​(ℓ)+Cl2​(g)→2BrCl(g) are 30kJmol−1 and 105JK−1mol−1 respectively. The temperature at which the reaction will be in equilibrium is :

285.7 K

373 K

250 K

400 K

For the reaction Br2​(ℓ)+Cl2​(g)→2BrCl(g) ΔH=30kJ/mol ΔS=105JK−1mol−1 For at equilibrium ΔG=0 ∴ΔG=ΔH−TΔS ΔH=TΔS T=ΔSΔH​=105JK−1mol−130×1000Jmol−1​ =2857K

Q. The enthalpy and entropy change for the reaction :
$B r_{2} (ℓ) + C l_{2} (g) \to 2 B r Cl (g)$

are $30 k J m o l^{- 1}$ and $105 J K^{- 1} m o l^{- 1}$ respectively. The temperature at which the reaction will be in equilibrium is :