The enthalpy and entropy change for the reaction textB ( textr)2 (. textl .) + textC ( textl)2 (. textg .) arrow 2 textB textr textC textl (. textg .) are 30 textk textJ text text textm texto textl - 1 texta textn textd 105 textJ textK - 1 text text textm texto textl - 1 respectively. The temperature at which the reaction will be in equilibrium is

Question

The enthalpy and entropy change for the reaction textB ( textr)2 (. textl .) + textC ( textl)2 (. textg .) arrow 2 textB textr textC textl (. textg .) are 30 textk textJ text text textm texto textl - 1 texta textn textd 105 textJ textK - 1 text text textm texto textl - 1 respectively. The temperature at which the reaction will be in equilibrium is (A) 285.7K (B) 273K (C) 450K (D) 300K

Tardigrade Admin · Accepted Answer

We know that text text textÎ textG = text text textÎ textH - textT textÎ textS When the reaction is in equilibrium, text text textÎ textG = 0 0= text text textÎ textH- textT textÎ textS⇒ textT=( textÎ textH/ text textÎ textS) textT=(30 × 1000/105)=285.7 textK

Q. The enthalpy and entropy change for the reaction
$B (r)_{2} (l) + C (l)_{2} (g) \to 2 B r C l (g)$
are $30 k J m o l^{- 1} a n d 105 J K^{- 1} m o l^{- 1}$ respectively. The temperature at which the reaction will be in equilibrium is

285.7K

273K

450K

300K

We know that
$Δ G = Δ H - T Δ S$
When the reaction is in equilibrium, $Δ G = 0$
$0 = Δ H - T Δ S \Rightarrow T = \frac{Δ H}{Δ S} T = \frac{30 \times 1000}{105} = 285.7 K$

Q. The enthalpy and entropy change for the reaction B(r)2​(l)+C(l)2​(g)→2BrCl(g) are 30kJmol−1and105JK−1mol−1 respectively. The temperature at which the reaction will be in equilibrium is