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nikitadnepr [17]
3 years ago
8

 A reaction takes place between two gases at 298 K. There are 2.0 moles of gas A and 3.0 moles of gas B. Which scenario would lo

wer the collision frequency between gases as well as the rate of reaction between gases A and B?A)Increase the moles of gas B.B)Lower the temperature of the reaction to 250 K.C)Compress the gases into a smaller volume.D)Introduce a catalyst to the container.
Chemistry
2 answers:
Free_Kalibri [48]3 years ago
7 0

Answer:

B)Lower the temperature of the reaction to 250 K.

Explanation:

Option B is correct. Lowering the temperature will decrease the kinetic energy of the gas molecules decreasing the frequency of collision and the rate of the reaction.

Option A is incorrect. Increasing the moles or the number of gas molecules of A and B will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option C is incorrect. Compressing the gas into a small volume will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option D is incorrect. A catalyst works by lowering the activation energy and thus increasing the frequency and orientation of the collision between gas molecules and thus increasing the rate of the reaction.  

Soloha48 [4]3 years ago
3 0
Choice B I think, because if the temperature decreases the particles move slower due to less kinetic energy, and thus won't collide as frequently.
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Answer:

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Explanation:

1. Calculate the moles of KCl needed

n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}

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3. Prepare the solution

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