Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
ap_chem:rate_of_reaction [2021/01/20 05:36] – epixap_chem:rate_of_reaction [2021/01/21 06:39] (current) – [Integrated Rate Laws] added epix
Line 6: Line 6:
  
 ===== 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|
-|No difference|Zeroth order| | +|No difference|0th order|$M^1*s^{-1}$| 
-|2x|First order| | +|2x|1st order|$M^0*s^{-1}$ or $s^{-1}$| 
-|4x|Second order| |+|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}}

Trace:

Navigation

bruh

QR Code
QR Code ap_chem:rate_of_reaction (generated for current page)