For the reaction 2 A( g) + B( g) ¡ 3 C( g),
b. when A is decreasi...

Question

For the reaction 2 A( g) + B( g) ¡ 3 C( g),
b. when A is decreasing at a rate of 0.100 M>s, how fast is
B decreasing? How fast is C increasing?

Accepted Answer

Hello. Everyone in this video. We're trying to consider the following hypothetical reaction. So that's this given reaction right over here being told that product Z. Is increasing at a rate of $0.240 per second. We're trying to determine the rate of disappearance of X. And the rate of appearance of Y. So let's go ahead and recall that if we have a reaction that is little A capital A. Yielding a little B, capital B. And the rate of the reaction is given to us by this equation. So that's negative one over a little A multiplied by the change of concentration of our capital A. Here divided by the change in times of delta T. And this is equal to one over little be multiplied by the chain of concentration of capital B. Or big B, divided by delta T. So again, we're given this reaction right over here. So then applying this kind of formula, the rate is then equal two and negative 1/3, multiplied by the concentration of X over delta T. This equals to 1/1. Multiplied by the concentration or the concentration of why over delta T. Which is equal to 1/2. Multiplied by the training concentration of capital Z, delta over delta T. So product Z is increasing at a rate of 0.240 molars per second. And that is the delta Z. Over delta teeth. Again this equal to 0.240 molars per second. So coupling for the rate of disappearance at X. Let's do this in red again. This is the rate for the disappearance. All right, so we have negative 1/3 multiplied by the change of concentration of delta X. Or the delta X. Which is the change in concentration of X. Over the delta T. Which is the change in temperature. This equals to 1/2. Multiplied by the change of concentration of Z. Divided by delta T. Alright, we're gonna go ahead and plug in the value that we do know. So that's on the right side. So on the left side we'll just copy everything as is so delta X. Over delta T. This equals to 1/2, multiplied by 0.240 molars per second. We're gonna go ahead and continue simplifying the right side of our equation, which gives us 0.1 to zero molars per second. We can isolate the delta X over delta T. We're gonna multiply both sides by negative three. If you do. So, we get that the delta X. Over delta T. Is unequal to negative 0. molars per second. So this is my rate of disappearance of X. So I'm gonna go ahead, highlight this for my first final answer. Then moving on to the rate of appearance of why. Alright, so here we have 1/1. Multiplied by the change of concentration of Y. Over delta T. So changing time, you're going to 1/2 by the chain of concentration of capital Z. Over delta T. So again, just isolating the change in concentration of Y over the change in time we get this is equal to 1/2, multiplied by 0.240 molars per second. Once we do, so we get that delta Y over delta T. Is unequal to $0.120 per second. So this is going to be my last and final answer for this problem. And this in green is just struggling for the rate of the appearance of why.

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