I need the calculation step
= 2. 0.5 mol solid benzene and 0.5mol liquid benzene reached equilibrium inside an adiabatic container, while 0.8mol ice and 0.2mol liquid water reached equilibrium inside another adiabatic container. It is known that the pressure inside both containers are the same at 101.325kPa. The freezing points of benzene and water are 5°C and 0°C, respectively. Also, the molar heat capacity of solid benzene under constant pressure is Cp,m (solid benzene) = 122.59) · K-1.mol-1, molar heat capacity of water under constant pressure is Cp,m (water) = 75.37 ): K-1. mol-4,the standard molar enthalpy of fusion for benzene is AH fus,m (benzene) = 9916) ·mol-1, the standard molar enthalpy of fusion for ice is ΔΗθ, fus,m (water) = 6004 ) · mol-1. a. The two systems are connected with a heat conductive layer and allow to reach equilibrium. Calculate the final temperature. (10 marks) = = . b. Calculate the total entropy change of this process. (10 marks) t = 3.878 °C = AS system = 0.3267) ·K-1
= 2. 0.5 mol solid benzene and 0.5mol liquid benzene reached equilibrium inside an adiabatic container, while 0.8mol ice
-
answerhappygod
- Site Admin
- Posts: 899604
- Joined: Mon Aug 02, 2021 8:13 am
= 2. 0.5 mol solid benzene and 0.5mol liquid benzene reached equilibrium inside an adiabatic container, while 0.8mol ice
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!