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

Chemistry Problems - Scholarship Essay ExampleAs the activation energy decreases the rate of reception increases. So the rate of the response would be fastest for the reaction with minimum activation energy (Ea = 10 kJ/mole).19. From the information it can be seen that as the parsimoniousness of Br2 is multiply keeping the concentration of NO constant, the rate of reaction is doubled. This gist that the reaction is first sight with prise to Br2.20. From the data it can be seen that as the concentration of reactant A is increase 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 in two ways the order 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 re actant A therefore the order of the reaction with respect to B is zero.2. According to the Le Chatliers principle, for a exothermic reaction if the temperature of the reaction is increased the value of the equaliser 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=1klnAtTherefore 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 devoted reaction 4 NH3 + 7 O2 4 NO2 + 6 H2OFrom the reaction i t 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 21. For the given reaction the equilibrium constant of the reaction can be given asK= HI2H2I2=0.5520.15*0.33=6.112. According to the Le Chatliers 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 reactionCO (g) + 3H2 (g) CH4 (g) + H2O (g)Q=CH4H2OCOH23=0.0620*0.05500.0450*0.1323=32.95Given that equilibrium constant K=3.93. So QK and therefore the reaction would proceed in the backward direction. 7. For the reaction6CO2 (g) + 6H2O (l) C6H12O6 (s) + 6O2 (g)The equilibrium constant would be given byK=C6H12O6O26CO26H2O6As the activity for the solid and liquid compounds can be taken to be 1, therefore the net equilibrium constant would be given byK=O26CO268. Fr om the given data the initial molarity of NOCl is