Answer to Question #183552 in General Chemistry for avani

Question

A 0.750 mole sample of HBr is introduced into an evacuated (empty) 2.00 L flask and kept at a temperature of 175°C. When the system comes to equilibrium the flask is found to contain 0.206 mole each of H₂ and Br₂. Determine the value of Kc at 175°C for the equilibrium reaction:

2HBr(g)  ↔ H₂(g) + Br₂(g)

Answer 1

80.91 × 0.750= 60.6825 × 2 = 121.365g

0.206 × 2.0157 = 0.415

= 0.206 × 159.808 = 32.92

K = H₂² (Br)/HBr²

= 0.206² × 0.206/0.75²

= 0.008742/0.5625

= 0.01554

= 1.554 ×10^-2