Question

The rate of disappearance of HBr in the gas phase reaction

2 HBr(g) → H2(g) + Br2(g)
is 0.221 M s^-1 at 150°C. The rate of reaction is ________ M s^-1.
a) 0.221
b) 0.442
c) 0.1105
d) 0.05525
e) none of these

Answer 1

The rate of reaction for the decomposition of 2 HBr(g) into H2(g) and Br2(g), given the rate of disappearance of HBr is 0.221 M s-1, is 0.1105 M s-1. the correct answer is C) 0.1105 M s-1.

 

Explanation:

The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.221 M s-1 at 150°C. To calculate the rate of reaction, we use the stoichiometry of the equation. Since 2 moles of HBr produce 1 mole of H2 and 1 mole of Br2, the rate of reaction will be half of the rate of disappearance of HBr.

The rate of a chemical reaction is typically expressed in terms of the change in concentration of a reactant or product per unit time. In the given gas-phase reaction .Since the balanced chemical equation indicates that two moles of HBr are consumed for every mole of the reaction, the rate of the overall reaction r) is half of the rate of disappearance of HBr: Therefore, the rate of reaction is 0.221 M s-1 ÷ 2 = 0.1105 M s-1. Therefore, the correct answer is 0.1105 M s-1.

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