At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:
H₂(g) + Br₂(g) 2HBr (g)
A mixture of 0.682 mol of H₂and 0.440 mol of Br₂is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.566 mol of H₂present. At equilibrium, there are
Mol of Br₂present in the reaction vessel.

Question

At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:
H₂(g) + Br₂(g)

At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H<sub>2 </sub>(g) + Br<sub>2</sub><sub> </sub>(g) 2HBr (g) A mixture of 0.682 mol of H<sub>2</sub><sub> </sub>and 0.440 mol of Br<sub>2</sub><sub> </sub>is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.566 mol of H<sub>2</sub><sub> </sub>present. At equilibrium, there are Mol of Br<sub>2 </sub>present in the reaction vessel. A) 0.440 B) 0.324 C) 0.000 D) 0.566 E) 0.232

2HBr (g)
A mixture of 0.682 mol of H₂and 0.440 mol of Br₂is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.566 mol of H₂present. At equilibrium, there are
Mol of Br₂present in the reaction vessel.

Quizplus · Accepted Answer

The Answer of At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H₂(g) + Br₂(g) 2HBr (g) A mixture of 0.682 mol of H₂and 0.440 mol of Br₂is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.566 mol of H₂present. At equilibrium, there are Mol of Br₂present in the reaction vessel.

At Elevated Temperatures, Molecular Hydrogen and Molecular Bromine React to Partially