The rate constant of the elementary reaction CH3OCH3(g) CH4(g) +CH2O(g) is k = 8.33×10-6 s-1 at 427°C, and the reaction has an activation energy of 245 kJ mol-1. (a) Compute the rate constant of the reaction at a temperature of 545°C. s-1 (b) At a temperature of 427°C, 8.32×104 s is required for half of the CH3OCH3 originally present to be consumed. How long will it take to consume half of the reactant if an identical experiment is performed at 545°C?
Answers
Answer:
(a) The rate constant is 3.61×10^-3 s^-1
(b) 7.12×10^4 s
Explanation:
(a) Log (K2/K1) = Ea/2.303R × [1/T1 - 1/T2]
K1 = 8.33×10^-6 s^-1
Ea = 245 kJ = 245,000 J
R = 8.314 J/mol.K
T1 = 427°C = 427+273 = 700 K
T2 = 545°C = 546+273 = 818 K
Log (K2/8.33×10^-6) = 245,000/2.303 × [1/700 - 1/818]
Log (K2/8.33×10^-6) = 2.637
K2/8.33×10^-6 = 10^2.637
K2 = 8.33×10^-6 × 433.51 = 3.61×10^-3 s^-1
(b) The relationship between temperature and the time required for reactants to be consumed is inverse
t2 = T1t1/T2
T1 = 427 °C = 700 K
t1 = 8.32×10^4 s
T2 = 545 °C = 818 K
t2 = 700×8.32×10^4/818 = 7.12×10^4 s
Related Questions
WORTH A LOT OF POINTS! just copy what on the picture for notes so i can copy and paste i do not feel like writing all of that down
An element is to a __ as an organ is to a ___
3Identify the ions present in K2HPO4.1) K+ and H+ and PO43-62) K 2+ H+, and PO43-93) K+, H+, p3-, and O2-4) K2H3+ and PO43-5) K2HPO4 is not ionic.
1. How do you determine how many protons are in a neutral element?
Molar mass of the weak base = 82.0343g/mole.
Note: pKa = -logKa
pKb = -logKb
pH + pOH = 14
[H+ ] [OH- ] = 10^-14
Answers
Answer:
11.39
Explanation:
Given that:
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
pH = 14 - pOH = 14 - 2.61 = 11.39
Answers
Answer: 75 liters of in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 Lat STP and contains avogadro's number of particles.
To calculate the number of moles, we use the equation:
According to stoichiometry:
1 mole of propane combines with = 5 moles of oxygen
Thus 0.67 moles of propane combine with =
Volume of
Thus 75 liters of in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.
conduction
convection
radiation
Answers
Fe2+, cr4+, cl-, O2-
Answers
Answer:
{eq}Fe^{2+} {/eq} and {eq}I^- {/eq} forms {eq}FeI_2 {/eq}
{eq}Fe^{2+} {/eq} and {eq}S^{2-} {/eq} forms {eq}FeS {/eq}
{eq}Cr^{4+} {/eq} and...
Explanation:
Empirical formula:
The empirical formula gives the simple ratio of the different types of atoms in a compound. It is different from the molecular formula, which gives the exact number of each type of atom in a compound.
Final answer:
In basic chemistry, an empirical formula represents the simplest ratio of atoms in a compound. For example, the empirical formula for a compound formed by Fe2+ (Iron II) and O2- (Oxide) would be FeO. In this question, a compound composed of all these ions (Fe2+, Cr4+, Cl-, O2-) is unusual and there's insufficient information to determine a reasonable structure.
Explanation:
The question is asking for an empirical formula, which is a formula that gives the simplest whole number ratio of atoms of each element in a compound. The formula you're asked to provide involves the ions Fe2+ (Iron II), Cr4+ (Chromium IV), Cl- (Chloride), and O2- (Oxide).
Creating an empirical formula is a matter of balancing out the charges in order to get a neutral compound. For instance, if you wanted to combine Fe2+ and O2-, the empirical formula would be FeO because one Fe2+ ion would balance out one O2- ion to make an electrically neutral compound.
It's important to remember that the charge value of the ion helps you determine the necessary ratio to achieve neutrality. In essence, we need the amount of positive charge to equal the amount of negative charge in the empirical formula.
For a compound involving all these ions, unfortunately, it's not common or reasonable to have a compound with four different ions. Iron, chromium, and oxygen are transition metals that could form complex ions, but we do not have enough information in this question to determine the structure.
Learn more about Empirical Formula here:
#SPJ2
Answers
Answers
Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.
The correct options are:
(A) - (a)
(B) - (d)
What are the most and the least acidic hydrogen atom?
The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.
The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.
Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.
Learn more about hydrogen atom, here:
The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.
The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.
Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.
Hope this helps!