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ap_chem:rate_of_reaction [2021/01/20 05:28] – created epixap_chem:rate_of_reaction [2021/01/21 06:39] (current) – [Integrated Rate Laws] added epix
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 ====== Rate of Reaction and Rate Law ====== ====== Rate of Reaction and Rate Law ======
 +
 For most equations you can write a rate law in the form of $r = k[X]^m[Y]^n$, where r is the rate of appearance of a product compound, $X$ and $Y$ being the reactant compounds, $k$ being the rate constant, and $m$ and $n$ being the order of the reactants. For most equations you can write a rate law in the form of $r = k[X]^m[Y]^n$, where r is the rate of appearance of a product compound, $X$ and $Y$ being the reactant compounds, $k$ being the rate constant, and $m$ and $n$ being the order of the reactants.
  
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 ===== Finding Orders ===== ===== Finding Orders =====
-{{: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.+{{: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. 
 + 
 +^Muliplier^Order^Unit| 
 +|No difference|0th order|$M^1*s^{-1}$| 
 +|2x|1st order|$M^0*s^{-1}$ or $s^{-1}$| 
 +|4x|2nd order|$M^{-1}*s^{-1}$| 
 +|8x|3rd order|$M^{-2}*s^{-1}$| 
 + 
 +As $H_2$ has a 2:2 difference, $H_2$ is a **1st order** reaction, consequently, because $NO$ is a 2:4 difference, it would be **2nd order**. Thus, the rate law for this reaction is $r = k[H_2][NO]^2$. 
 + 
 +===== Solving for $k$ ===== 
 +We can use one of the table entries to solve for $k$ using the rate law we just created. The rate of production would be $r$ and the values for $NO$ and $H_2$ would be placed in their respective slots. 
 + 
 +  - $r = k[H_2][NO]^2$ 
 +  - $1.8 \times 10^{-4} = k[0.001][0.006]^2$((when solving treat the brackets as parentheses)) 
 +  - $\frac{1.8 \times 10^{-4}}{3.6 \times 10^{-8}} = k$ 
 +  - $k = 5000$ $M^{-2}*s^{-1}$ 
 + 
 +The unit for each order is given to the right of the previously mentioned table, and the sum of the rate law's orders corresponding unit would be the $k$ value's unit. 
 + 
 +===== Solving Rate of Production or $r$ ===== 
 +Now that we have an equation and $k$ value, we can estimate rates of production of $N_2$ by plugging and chugging values into our new rate law. Take trial 5 for example: 
 + 
 +  - $r = k[H_2][NO]^2$ 
 +  - $r = 5000[0.002][0.002]^2$ 
 +  - $r = 4 \times 10^{-5}$ $\frac{\text{mol}}{L}*\text{min}$ 
 + 
 +===== Rate of Disappearance ===== 
 +Rate of Disappearance is similar to Production, after solving for the production of $N_2$, use [[ap_chem:stoichiometry|stoichiometry]] to find the rate of disappearance. Looking back at our equation, there is a 2:1 ratio of $NO$ to $N_2$, so we'll use that. 
 + 
 +$ 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.
  
 +===== 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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