<p>I have summer work for AP Chem and was hoping to see if someone will help me with some problems.</p>
<li><p>A reaction has a rate constant of 6.3*10^4 L^2 mol^-2 s^-1. One possible rate law for this reaction is:
A) Rate = k[A]^2**
B) Rate = k[A]^2
C) Rate = k[A]^2
D) Rate = k[A]
E) Rate = k^-1</p></li>
<li><p>When the following half-reaction:
CrO4^2- → Cr^3+
is balanced with the lowest possible coefficients in acid solution, there will be
A) three electrons on the right
B) six electrons on the right
C) no electrons needed
D) three electrons on the left
E) two electrons on the right</p></li>
<li><p>N2O4 dissociates into two NO2 molecules. When 0.500 mole of N2O4 is placed in a 2.00-liter flask and allowed to come to equilibrium, the final concentration of N2O4 is 0.100 mole per liter. What is the equilibrium constant for the dissociation of N2O4 under these conditions?
A) 0.200
B) 0.900
C) 0.225
D) 4.44
E) 1.11</p></li>
</ol>
<p>Kinda guessed, the units look like second order
3. B
N2O4 <> 2NO2 so K = [NO2]^2 / [N2O4]</p>
<pre><code>N2O4 <> 2NO2
</code></pre>
<p>I .25 M 0 M
C -.15 M +2(.15M)
E .1 M .30M</p>
<p>K= (.30)^2 / .10</p>
<p>Was this given by your teacher? This was like third quarter material for my class this past year. Ideal summer work should be stoic review unless you are self-studying.</p>
<p>for 1: A or C, since when u multiply it out u have to get mol s^-1
2: D
Cr^6+ in CrO4, so 3 e- added to Cr)4 i think. i dont kno how to balance it when there is no O2 on the LHS
3: I agree w/ snipez</p>