(a)
Interpretation:
What are the units of constants 200 and 10.
Concept introduction:
Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as,
It is assumed that the given equation is valid and therefore dimensionally homogeneous. So, the dimension of the left hand side and right hand side of the equation is equal. In addition, the dimensions in the two sections of the left hand side are equal.
(b)
Interpretation:
The crystal nucleation rate to a crystal diameter of 0.050 inch should be calculated.
Concept introduction:
Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as,
(c)
Interpretation:
Derive a formula for r (crystals/s) in terms of D (inches).
Concept introduction:
Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as,
(d)
Interpretation:
Explain what is the equation gives away with its empirical nature.
Concept introduction:
Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as,
It is assumed that the given equation is valid and therefore dimensionally homogeneous. So, the dimension of the left hand side and right hand side of the equation is equal. In addition, the dimensions in the two sections of the left hand side are equal.
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EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
- The plot below shows the concentration of reactant A during a chemical reaction (A (ag) + B(ag) C (aq) + D (aq)) as a function of time. This reaction is taking place at 310 K A and B are aqueous solutes Estimate the instantaneous rate at 80 seconds. Give your answer with units of M/s. 0.50 (A) (M) 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 005 0.00 0 20 Type your answer. 40 60 time (s) 80 100 120 140arrow_forwardYou are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed a competitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Km of the enzyme was found to be 1.5 x 10-5 M. What is the Ki of the inhibitor?arrow_forwardIn a study of the conversion of methyl isonitrile to acetonitrile in the gas phase at 250 CH 3 NC(g) CH 3 CN(g) the concentration of CH 3 NC was followed as a function of time It was found that a graph of ln[CH 3 NC] versus time in seconds gave a straight line with a slope of - 4.17 * 10 ^ - 3 * s ^ - 1 and a y intercept of -4.05 Based on this plot , the reaction is order in CH 3 NC and the rate constant for the reaction isarrow_forward
- Nitric oxide reacts with hydrogen to release large amounts of chemiluminescence with the characteristics of the highly cytotoxic species. The following data were measured for the reaction of nitric oxide with hydrogen: Data number 1 2 3 2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g) [NO] (M) 0.1 0.1 0.2 [H₂] (M) 0.1 0.2 0.1 Based on the data, (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H₂] = 0.150 M Initial rate (M/S) 1.23 x 10-³ 2.46 x 10-³ 4.92 x 10-³arrow_forwardЗа. Bromine and methanoic acid react in aqueous solution. Br, (aq) + HCO0Н (ад) — 2Br (aq) + 2H" (аq) + Со, (g) The reaction was monitored by measuring the volume of carbon dioxide produced as time progressed. o 5 10 15 20 25 [Source: © International Baccalaureate Organization 2019] 30 20 40 60 80 100 120 140 160 Time / s [Source: © International Baccalaureate Organization 2019] Determine from the graph the rate of reaction at 20 s, in cm³ s1, showing your working. 3b. Outline, with a reason, another property that could be monitored to measure the rate of this reaction. 3c. Describe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate. 3d. Identify one error associated with the use of an accurate stopwatch. Volume of carbon dioxide / cm T m mm moarrow_forward1-arr In a study of the gas phase decomposition of phosphine at 120 °C PH3(g)- →1/4 P4(g) + 3/2 H2(g) the concentration of PH3 was followed as a function of time. It was found that a graph of In[PH3] versus time in seconds gave a straight line with a slope of -2.50×10² s1 and a y-intercept of -3.24. Based on this plot, the reaction i v O order in PH, and the half life for the reaction is seconds. zero first secondarrow_forward
- Time(s) [X3] (mole/L) 0 0.600 500 0.481 1000 0.392 1500 0.336 2000 0.273 2500 0.230 3000 0.181 3500 0.156 4000 0.120 4500 0.106 5000 0.072 plot ln[X3] vs time plot 1/[X3] vs. timearrow_forwardAn environmental chemist working for the Environmental Protection Agency (EPA) was directed to collect razor clams from a heavily-contaminated river superfund site and analyze them for their Cd?+ content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist dried the clams at 95 °C overnight and ground them in a scientific blender, resulting in approximately 50 g of homogenized dry weight. A representative 98.75 mg sample was taken from the approximately 50 g of dry material and dissolved in 100.0 mL of 0.1 M HCI to create a sample solution. Using the method of standard additions, the chemist prepared five standard solutions in 100.0 mL volumetric flasks, each containing 5.00 mL aliquots of the sample solution. Varying amounts of a 75.0 ppb (ug/L) Cd? + standard were added to each of the flasks, which were then brought to volume with 0.1 M HCI. The Cd? + content of the solutions was then analyzed using GFAAS, resulting in the. absorbance data given in the table.…arrow_forward2. a) Calculate the hard-sphere (HS) collision theory rate constant ks for the reaction NO(g) + C₁₂(g) Þ NOCl(g) + HS Cl(g) at 300 K in units of dm³ mols. The collision diameters of NO and Cl₂ are 370 and 540 pm, respectively. 16.0 g/mol, and mcı Use masses of m₁ = 14.0 g/mol, mo = 35.0 g/mol. N = b) Now include the line-of-centers (LOC) energy criterion to find koc in units of dm³ mol¹ s¹ using E₁ = 84.9 kJ mol LOC LOC c) Calculate the ratio of kчs and kLoc to the experimental rate constant at 300 K given A = 3.981 10° dm³ mol¹ s and E₁ = 84.9 kJ mol¹. a -1arrow_forward
- 1-arr In a study of the decomposition of dinitrogen pentoxide in carbon tetrachloride solution at 30 °C N2O5- →2 NO2 + ½ O2 the concentration of N,O5 was followed as a function of time. It was found that a graph of In[N2O5] versus time in minutes gave a straight line with a slope of -6.31×10* min and a y- intercept of -1.34 . Based on this plot, the reaction O order in N,0g and the rate constant for the reaction is min-1. zero first secondarrow_forward5R) The net ionic equation for formation of an aqueous solution of Nil2 accompanied by evolution of CO2 gas via mixing solid NICO3 and aqueous hydriodic acid is A) 2NICO3 (s) + HI (aq) - 2H20 (1) + CO2 (g) + 2N12+ (aq) 2H20 (1) + CO2 (g) + Ni2+ (aq) + HI (aq) B) NICO3 (s) + I (aq) O NICO3 (s) + 2H* (aq) H20 (1) + CO2 (g) + Ni2+ (aq) 2H20 (1) + CO2 (g) + Nil2 (aq) D) NICO3 (s) + 2HI (aq) E) NICO3 (s) + 2HI (aq) H20 (1) + CO2 (g) + Ni2+ (aq) + 21- (aq) Answer. Carrow_forwardTrial [I0:] [HSO: ] Time Rate [HSO3 ]/time 1 49 25 3 16.9 Table-2 Find the concentrations of [IO3] and [HSO3] and enter into the table. Divide the concentration of HSO3 with the time in seconds to get the rate. Volume is assumed to be additive in this experiment. Thus, the total volume of all the solutins is assumed to be 20.0 mL. This introduces a small error that can be considered insignicant. The concentrations of IO; and HSO, (after mixing but before the reaction begins) can be calculated using the relationship: Mailute Vdilute= Mcone Vcone. A sample calculation is shown below for the concentration of IO, in Trial 1: Mailute to be calculated Vdilue 20.0 mL 0.020 M x 2.0 mL Mcone 0.020M V conc 2.0 mL %3D diluted [IO, ] = = 0.0020 M IO, %3D 20.0 mL Similarly, the concentration of HSO; in Trial 1 is calculated thus: 0.010 M x 2.0 mL diluted [HSO, ] =0.0010 M HSO, 20.0 mLarrow_forward
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