Chemistry Problems Scholarship Essay Example | Topics and Well Written Essays - 1250 words

Chemistry Problems - Scholarship Essay Example As the activation energy decreases the rate of reaction increases. So the rate of the reaction would be fastest for the reaction with minimum activation energy (Ea = 10 kJ/mole). 19. From the data it can be seen that as the concentration of Br2 is doubled keeping the concentration of NO constant, the rate of reaction is doubled. This means that the reaction is first order with respect to Br2. 20. From the data it can be seen that as the concentration of reactant A is increased to 4 times, keeping the concentration of B and C constant, the rate of the reaction increases by 4 times. This means that the reaction is first order with respect to A. Similarly when the concentration of reactant A is increased to twice the value and that of reactant B is also increased to twice the value, keeping the concentration of reactant C constant, the rate of reaction increases to twice the value. As this increment is attributed to reactant A therefore the order of the reaction with respect to B is zero. 2. According to the Le Chatlier's principle, for a exothermic reaction if the temperature of the reaction is increased the value of the equilibrium constant decreases. As the equilibrium constant decreases the concentration of products is lowered. .. This means that the reaction is first order with respect to A. Similarly when the concentration of reactant A is increased to twice the value and that of reactant B is also increased to twice the value, keeping the concentration of reactant C constant, the rate of reaction increases to twice the value. As this increment is attributed to reactant A therefore the order of the reaction with respect to B is zero. 23. We know that for a first order reaction: t=1klnAt Therefore for a first order reaction, the plot of At versus t would be a logarithmic curve and not a straight line. 24. For the stoichiometry of the given reaction: 4 NH3 + 7 O2 4 NO2 + 6 H2O From the reaction it can be seen that highest moles of O2 are consumed in the reaction. Therefore it can be assumed that the O2 is removed the fastest during the reaction. Module 2 1. For the given reaction the equilibrium constant of the reaction can be given as: K= [HI]2H2I2=0.5520.15*0.33=6.11 2. According to the Le Chatlier's principle, for a exothermic reaction if the temperature of the reaction is increased the value of the equilibrium constant decreases. As the equilibrium constant decreases the concentration of products is lowered. 4. For the reaction: CO (g) + 3H2 (g) CH4 (g) + H2O (g) Q=CH4H2OCOH23=0.0620*0.05500.0450*0.1323=32.95 Given that equilibrium constant K=3.93. So Q>K and therefore the reaction would proceed in the backward direction. 7. For the reaction: 6CO2 (g) + 6H2O (l) C6H12O6 (s) + 6O2 (g) The equilibrium constant would be given by: K=C6H12O6O26CO26H2O6 As the activity for the solid and liquid compounds can be taken to be 1, therefore the net equilibrium constant would be given by: K=O26CO26 8. From the given data the initial molarity of NOCl is