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Chemistry 1202 Final Exam for John Hogan practice one

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Chemistry 1202 Final Exam for John Hogan Practice One 1. 2. For the gas-phase reaction shown below, choose the answer which indicates how the rate of change of N2 is related to the rate of change of O... 2. 4 NH3(g) + 3 O2(g) 2 N2(g) + 6 H2O(g) a) - 1/2 [N2] /t = - 1/3 [O2] /t b) 1/2 [N2] /t = 1/3 [O2] /t c) 1/2 [N2] /t = - 1/3 [O2] /t d) 1/3 [N2] /t = - 1/2 [O2] /t e) 3 [N2] /t = 2 [O2] /t (-1/Coefficient for O2) and (1/Coefficient for N2) 3. Consider the reaction of tetrafluorohydrazine with hydrogen: 1 N2F4(g) + 5 H2(g) 2 NH3(g) + 4 HF(g). If NH3 is produced at the rate of 0.29 mol/s, what is the rate of production of HF? a) The rate of production of HF is 2.3 mol/s. b) The rate of production of HF is -0.58 mol/s. c) The rate of production of HF is 0.15 mol/s. d) The rate of production of HF is 0.58 mol/s. e) The rate of production of HF is -0.15 mol/s. (0.29 mol/s NH3) * (4 mol HF/ 2 mol NH3) = 0.58 mols/s HF 4. Consider the following reaction: 2 HgCl2(aq) + C2O42-(aq) 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s) The rate law for this reaction is first order in HgCl2(aq) and second order in C2O42-(aq). Pick the correct rate law for this reaction from the multiple choices. a) Rate = k [HgCl2(aq)] [C2O42-(aq)] b) Rate = k [HgCl2(aq)] [C2O42-(aq)]2 c) Rate = k [HgCl2(aq)] [C2O42-(aq)]4 d) Rate = k [HgCl2(aq)]2 [C2O42-(aq)] e) Rate = k [HgCl2(aq)]4 [C2O42-(aq)] First Order = X^1 Second Order = X^2 and so on 5. Consider the following reaction: 2 ClO2(aq) + 2 OH-(aq) ClO3-(aq) + ClO2-(aq) + H2O(l) The rate law for this reaction is: Rate = k [ClO2(aq)]2 [OH-(aq)] If the rate constant for this reaction at a certain temperature is 4.05e+02 M-2 s-1, what is the reaction rate when [ClO2(aq)] = 0.0509 M and [OH-(aq)] = 0.0373 M? a) The reaction rate is 3.91e-02 M s-1. b) The reaction rate is 2.39e-07 M s-1. c) The reaction rate is 5.83e+03 M s-1. d) The reaction rate is 2.56e+01 M s-1. e) The reaction rate is 2.81e+01 M s-1. Use the given rate law and plug in the numbers Rate = (4.05e+2) * (0.0509)^2 *(0.0373) [Show More]

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