Question
Given that bond energies of H - H and Cl - Cl are 430 ~kJ ~mol ^ -1 and 240 ~kJ ~mol ^ -1 respectively and H _ f for HCl is -90 ~kJ ~mol ^ -1 , bond enthalpy of HCl is
BITSAT
Chemistry
Thermodynamics (C)
2023
BITSAT 2023 (Memory Based Paper 2)
BITSAT Chemistry Question (2023) — Solution
Options
- A. 380 ~kJ ~mol ^ -1
- B. 425 ~kJ ~mol ^ -1
- C. 245 ~kJ ~mol ^ -1
- D. 290 ~kJ ~mol ^ -1
Answer
B. 425 ~kJ ~mol ^ -1
Step-by-step solution
1 2 H _2+ 1 2 Cl _2 HCl aligned & H_ HCl = B.E. of reactant - B.E. of products \\ &-90= 1 2 430+ 1 2 240- B.E. of HCl \\ & B.E. of HCl =215+120+90=425 ~kJ ~mol ^ -1 aligned
Related: Chemistry — Thermodynamics (C) · All PYQ Banks