<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B79%5EoC%7D%7D%7BK_%7B26%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 79°C = 
= equilibrium constant at 26°C = 
= Activation energy of the reaction = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![26^oC=[26+273]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B26%2B273%5DK%3D299K)
= final temperature = ![79^oC=[79+273]K=352K](https://tex.z-dn.net/?f=79%5EoC%3D%5B79%2B273%5DK%3D352K)
Putting values in above equation, we get:
![\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B0.394%7D%7B2.08%5Ctimes%2010%5E%7B-4%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B299%7D-%5Cfrac%7B1%7D%7B352%7D%5D%5C%5C%5C%5CE_a%3D124595J%2Fmol%3D124.6kJ%2Fmol)
Hence, the activation energy of the reaction is 124.6 kJ/mol
This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:
I. C1=CC=CC=C1C(=O)C
II. C1=CC=CC=C1CC=O
III. C1=CC(C)=CC=C1C=O
IV. C1=CC=CC=C1CCC
V. C1=CC=CC=C1C(C)C
The structures are also attached
Answer:
The structure of compound IV is consistent with the information obtained analysis
Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):
C1=CC=CC=C1C(=[OH0+])C |^1:7|
C1C([CH0+]=O)=CC=CC=1
C1[CH0+]=CC=CC=1
C(#[OH0+])C
respectively
Explanation:
The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.
Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)
The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
Air is considered a homogeneous mixture. Hope that helps :)
Answer:
a. H2S(g)/t = 1.48 mol/s
CS2(g)/t = 0.740mol/s
H2(g)/t = 2.96mol/s
b.
Ptot /t = 981torr/min
Explanation:
a. Based on the reaction:
CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)
<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>
<em />
If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s
CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s
The H2 is produced four times faster than CH4 is decreased, that is:
0.740mol/s * 4 = 2.96mol/s
b. With the reaction:
2 NH3(g) → N2(g) + 3 H2(g)
2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.
That means 2 moles of gas are consumed and 4 moles of gas are produced.
If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =
654torr/min
The rate of change of the total pressure is rate of reactants + rate of products:
654torr/min + 327torr/min =
981torr/min
