The rate of the reaction?

PLEASE HELP A reaction has the rate law: Rate = k[A]^2[Z]^3/2. If the concentration of Z is increased by a fac?

• Please help I have the answers to these but I need to know how to work these out. I have a test on Monday and the questions are going to be similar to these. Please show all the work and steps so I can learn and not be lost during the test. Thanks! 1) A reaction has the rate law: Rate = k[A]^2[Z]^3/2. If the concentration of Z is increased by a factor of four, the reaction rate will increase by a factor of _____. Answer: 8 2) The average rate of disappearance of ozone in the reaction 2O3(g) --->3O2(g) is found to be 8.36x10^-3 atm over a certain interval of time. What is the rate of appearance of O2 during this interval? Answer: 12.5x10-^3 atm 3)Consider the reaction: 2NH3(g) ---> N2(g) + 3H2(g) If the rate, Δ[H2]/Δt is 0.030 mol/L∙ s, then Δ[NH3]/Δt is Answer: −0.020 mol/L∙s 4)Consider the reaction 2H2 + O2 ---> 2H2O What is the ratio of the initial rate of the appearance of water to the initial rate of disappearance of oxygen? Answer: 2:1 5) The reaction A--->B + Cis known to be zero order in A with a rate constant of 5.0x10^-2 mol/L s at 25°C. An experiment was run at 25°C where [A]o = 1.5x10^-3 M. The half-life (t1/2) for the reaction is Answer: 1.5x10^-2 s

• Answer:

  I can help with some. 1) This kind of question can be worked out by substituting real numbers. eg, 1^3/2= 1. You increase it by a factor of four, so multiply by four. 4^3/2=8. When Z increased by a factor of 4, the rate increased by a factor of 8. This works because whilst you are increasing the concentration of Z, the concentration of everything else is staying the same and will have no effect on the increase. 2) the equation occurs in a ratio of 2:3. divide 8.36x10^-3 by 2 (the ozone), which will tell you the rate for 1 mole of a substance and times by 3, which is the rate for O2, as it occurs, in this equation, with 3 moles. This gives you your answer. 3) like the question above and below, it's a matter of mole ratio: NH3 and H2 are, within this equation, occurring in a ratio of 2:3. H2, which is the 3, has a rate of 0.030mol/L.s . to get it down to one, divide by 3, and then times by 2 to get it to the amount NH3 occurs. 0.020mol/L.s. Because it is on the other side of the equation, it is a negative. 4) The number before the elements/compounds indicate the ratio the experiment is occurring in: 2 hydrogen molecules plus one oxygen molecule form two water molecules. The question asks the ratio of the water produced to oxygen lost, so that will be 2:1. I don't think you typed 5 out properly.