Reaction between N2 and O2– takes place as follows:
2N2 (g) + O2 (g) ↔ 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
Let the concentration of N2O at equilibrium be x. The given reaction is:
2N2 (g) + O2 (g) ↔ 2N2O (g)
Intial Concentration: 0.482 mol 0.933 mol 0 mol
At equilibrium (0.482-x) mol (0.933 -x) mol x mol
Therefore, at equilibrium, in the 10 L vessel:
N2 = 0.482-x / 10
O2 = 0.933-x/2 / 10
N2O = x / 10
The value of equilibrium constant i.e.Kc = 2.0 × 10-37 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.
Then,
N2 = 0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1
Now,
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
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(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
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(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
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(i) 1 g Au (s)
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(iii) 1 g Li (s)
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For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
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Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?
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(iii) < 0
(iv) different for each element
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