At 373 K steam and water are in equilibrium and D H = 40.98 kJ mol –1 . What will be D S…
Chemistry · Class 11 · CBSE — Thermodynamics
At 373 K steam and water are in equilibrium and DH = 40.98 kJ mol–1. What will be DS for conversion of water into steam?
H2O(l) → H2O(g)
H2O(l) → H2O(g)
- 109.8 J K–1 mol–1
- 31 J K–1 mol–1
- 21.98 J K–1 mol–1
- 326 J K–1 mol–1
Answer
(A) 109.8 J K –1 mol –1
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- Δ H for the reaction, OF 2 + H 2 O → O 2 + 2HF ( B . E . of O F, O H, H F and O O are 44, 111, 135 and 119…
- For a reaction : C ( s ) + O 2( g ) → CO 2( g ) What is the relation between enthalpy of reaction (Δ H R )…
- Bond energies of few bonds are given below : Cl Cl = 242.8 kJ mol –1 , H Cl = 431.8 kJ mol –1 , O H = 464 kJ…
- Formation of ammonia is shown by the reaction, N 2( g ) + 3H 2( g ) → 2NH 3( g ) , D r H ° = –91.8 kJ mol –1…
- The molar heat capacity of water at constant pressure, C P is 75 J K –1 mol –1 . When 10 kJ of heat is…
- For the reaction : H 2( g ) + Cl 2( g ) → 2HCl; D H = – 44 kcal What is the enthalpy of decomposition of HCl?
- For reversible reaction : X ( g ) + 3 Y ( g ) 2 Z ( g ) ; Δ H = – 40 kJ Standard entropies of X , Y and Z are…
- What will be the enthalpy change of conversion of graphite into diamond? Given C graphite , D comb H =…
More Thermodynamics questions · Browse all practice questions