Kinetics Chemistry Question [reaction mechanisms]?
954211
Your first conclusion is correct, but your second isn't.  Since you already have the mechanism given to you, you can see that the 2nd step is the slow step.  So, the rate law should be:\n\nrate = k (N2O2) (H2)\n\nSince N2O2 is an intermediate, it can't appear in the overall rate law, so you have to get rid of that.  From the first reaction, you see that N2O2 is formed from the collision of 2 NO.  So, the concentration of N2O2 is proportional to [NO]^2.  That gives you the overall rate law that you had, and the overall order of the reaction is 3rd order.\n\nYour second conclusion is incorrect because the slow step requires the collision of N2O2 and H2.  The slow step is just a second order reaction.  \n\nSo, you've got things pretty well in hand...Hope this helps a little...
857176
S(s) + O2(g)------>SO2\n\n2Mg(s) + O2(g) -------> 2MgO(s)\n\nH2(g) + Br2 (g) -------> 2HBr(g)\n\n4Fe(s) + 3O2 -------> 2Fe2O3 (s)\n\nThen there are single replacement \n\nA + BC-------> AC + B\nZn(s) + 2HCl(aq)------->ZnCl2 (aq) + H2 (g)\nwhere one element replaces another\n\nthen there is double replacement \n\nAB+CD -------> AD + BC\n\nIn this one the positive ions in the reacting compounds switch places\n\nBaCl2 (aq) + Na2So4 (aq) --------.BaSo4 (s) + 2NaCl (aq)
460906
DH reaction = DH products – DH reactants\n\nExample:  2A + B -- C\nDH = DH C – (2 DH A + DH B)\n\nIf DH is negative: exothermic reaction \nIf DH is positive: endothermic reaction
677341
The reaction is\n\nH2O2(aq) + 3I-(aq) + 2 H+(aq) I3-(aq) + 2 H2O(aq)\n\nThe reaction above is often very slow, so a catalyst is usually added. Ammonium molybdate works; so do other transition metal ions. The catalytic action probably involves a free radical mechanism. Hydrogen peroxide can reduce MnO4-, but molybdate isn't a strong enough oxidizing agent to be reduced.
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
444111
Rate of reaction depends upon the concentration of reactants. Since, 6M HCl has a greater conc. than 3M HCl, rate of reaction decreases when Mg reacts with 3M HCl
553744
http://en.wikipedia.org/wiki/Reaction_mechanism\nhttp://en.wikipedia.org/wiki/Bromomethane
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
284573
good link below\n\nhttp://en.wikipedia.org/wiki/Kinetic_theory
309732
i.e. it is easier to go backwards than forwards, so the product has more energy ( i.e. higher up the energy scale / nearer the top/ easier to get to the transition state ).\n\nHeat of reaction = heats of formation of products - reactants = 85 - 80 = +5kJ mol.... endothermic ( makes sense if products contain more energy than reactants )
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
971455
there are various factors which affects rate of reaction. like temperature,pressure and concentration of the reactants involved. and they all have different effects
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
418418
Hopefully this diagram can answer your questions:\nhttp://fachschaft.cup.uni-muenchen.de/~schleifi/reaktion/reaction/stobbe.html