Stoichiometric Rate Calculations Concept 1

Jules Bruno
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now start Geometric rate calculations happen when the rate of one compound is known and the rate of another can be calculated using a rate to read comparison. So here in this example, it says if the rate of decomposition of H2 is 1.54 polarities per minute at a particular time, What would be the rate of formation of nitrogen gas at that same time? Alright, so they're giving us the rate of one compound and asking for the rate of another within a balanced chemical equation. This is a key giveaway that you're dealing with a stark geometric rate calculation. Now here. Step one. If the rate of change for one compound is not given, then first calculated using information provided here. We don't have to worry about that because we're given the rate of H2 right from the beginning. It's 1.54. More clarity is per minute Now. Step two, using the rate of one compound perform, a rate to rate comparison using stark geometric coefficients. Now, this is similar to a multiple comparison that we've used before in stock geometry. Now the way we set it up is We have 1.54 more polarity per minute, and this is for H two. And remember, in stock geometry would do a multiple comparison. But now we're doing a rate to rate comparison. So we're gonna say, according to our balanced chemical equation, We have two year 4 H two. That, too, would mean to mola Garrity's per minute for each two And then for end to. There's a one here. So that's one Molinari teas per minute for end to. So here are units will cancel out. And look, we're going to have the rate of N two. So when we plug that in, we get 20.770 molar. It is per minute for N two, so this would be the rate of nitrogen gas. So again, it's kind of reminiscent a little bit of stock geometry. But instead of doing a multiple comparison, we're doing the rate to rate comparison, so approach it in the same manner and you'll always be able to find the rate of any compound within a balanced equation if they give you a rate of another compound within that same balanced equation,
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