The equilibrium constant for the reaction H 2( g )+ S ( s ) leftharpoons H 2 S ( g ); is 18.5 at 925 K and 9.25 at 1000 K respectively. The enthalpy of the reaction will be :

Question

The equilibrium constant for the reaction H 2( g )+ S ( s ) leftharpoons H 2 S ( g ); is 18.5 at 925 K and 9.25 at 1000 K respectively. The enthalpy of the reaction will be : (A) -68000.05 J mol -1 (B) -71080.57 J mol -1 (C) - 80071.75 J mol -1 (D) 57080.75 J mol -1

Tardigrade Admin · Accepted Answer

H 2( g )+ S ( s ) leftharpoons H 2 S ( g ) log ( K 2/ K 1)=(Δ H /2.303 R )((1/ T 1)-(1/ T 2)) log ((9.25/18.5)=(1/2))=(Δ H /2.303 × 8.314)((1/925)-(1/1000)) -0.3010=(Δ H /2.30 × 8.314)((75/925 × 1000)) Δ H =-71080.57 J / mole

Q. The equilibrium constant for the reaction $H_{2} (g) + S (s) ⇌ H_{2} S (g)$ ; is $18.5$ at $925 K$ and $9.25$ at $1000 K$ respectively. The enthalpy of the reaction will be :

$- 68000.05 J m o l^{- 1}$

$- 71080.57 J m o l^{- 1}$

$- 80071.75 J m o l^{- 1}$

$57080.75 J m o l^{- 1}$

Q. The equilibrium constant for the reaction H2​(g)+S(s)⇌H2​S(g); is 18.5 at 925K and 9.25 at 1000K respectively. The enthalpy of the reaction will be :

−68000.05Jmol−1

−71080.57Jmol−1

−80071.75Jmol−1

57080.75Jmol−1

H2​(g)+S(s)⇌H2​S(g) logK1​K2​​=2.303RΔH​(T1​1​−T2​1​) log(18.59.25​=21​)=2.303×8.314ΔH​(9251​−10001​) −0.3010=2.30×8.314ΔH​(925×100075​) ΔH=−71080.57J/mole

Q. The equilibrium constant for the reaction $H_{2} (g) + S (s) ⇌ H_{2} S (g)$ ; is $18.5$ at $925 K$ and $9.25$ at $1000 K$ respectively. The enthalpy of the reaction will be :

$- 68000.05 J m o l^{- 1}$

$- 71080.57 J m o l^{- 1}$

$- 80071.75 J m o l^{- 1}$

$57080.75 J m o l^{- 1}$