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ap_chem:rate_of_reaction [2021/01/20 06:01] – add note on rate of disappearance epixap_chem:rate_of_reaction [2021/01/21 06:39] (current) – [Integrated Rate Laws] added epix
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 ===== Finding Orders ===== ===== Finding Orders =====
 +{{:ap_chem:2021-01-19-232308_355x199_scrot.png?400  }}
  
-{{:ap_chem:2021-01-19-232308_355x199_scrot.png?400  }}Take the example diagram here for the reaction $2NO_{(g)} + 2H_{2 (g)} -> N_{2 (g)} + 2H_2O_{(g)}$. To determine $m$ and $n$ for the rate law, you have to find pairs of trials that only modify **1** of the reactant compound's quantity, divide it, and compare the quotient to the reaction rate's quotient((as seen in the blue and green markup)) . Refer to the table below for the corresponding order.+Take the example diagram here for the reaction $2NO_{(g)} + 2H_{2 (g)} -> N_{2 (g)} + 2H_2O_{(g)}$. To determine $m$ and $n$ for the rate law, you have to find pairs of trials that only modify **1** of the reactant compound's quantity, divide it, and compare the quotient to the reaction rate's quotient((as seen in the blue and green markup)) . Refer to the table below for the corresponding order.
  
 ^Muliplier^Order^Unit| ^Muliplier^Order^Unit|
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 $ 4 \times 10^{-5} * \frac{2NO}{1N_2} = -8 \times 10^{-5}$ $\frac{\text{mol}}{L}*\text{min}$((the negative is there because it is a rate of disappearance!)) which is our rate of disappearance of $NO$ for trial 5. $ 4 \times 10^{-5} * \frac{2NO}{1N_2} = -8 \times 10^{-5}$ $\frac{\text{mol}}{L}*\text{min}$((the negative is there because it is a rate of disappearance!)) which is our rate of disappearance of $NO$ for trial 5.
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 +===== Integrated Rate Laws =====
 +Use the below chart to determine for each order. Note that half life problems on the AP test will always be given with first order for simplicity.
 +
 +{{:ap_chem:5ae8d514bba2df4f1a350329d94c3d8b.jpg?800}}

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