Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ↔ 2NO (g) + Cl2 (g); Kp = 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ↔ CaO(s) + CO2(g); Kp = 167 at 1073 K
(a) The relation between Kp and Kc is given as:
Kp = Kc (RT)Δn
(a) Here, Δn = 3 - 2 = 1
R = 0.0831 barLmol-1K-1
T = 500 K
Kp = 1.8 x 10-2
Now, Kp = Kc (RT)Δn
⇒ Kc = 1.8 x 10-2 / (0.0831x500)
= 4.33 x 10-4 (approx.)
(b) Here, Δn = 2 - 1 = 1
R = 0.0831 barLmol-1K-1
T = 1073 K
Kp = 167
Now, Kp = Kc (RT)Δn
⇒ Kc = 167 / (0.0831x1073)
= 1.87 (approx.)
The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements | ΔiH1 | ΔiH | ΔegH |
I | 520 | 7300 | -60 |
II | 419 | 3051 | -48 |
III | 1681 | 3374 | -328 |
IV | 1008 | 1846 | -295 |
V | 2372 | 5251 | +48 |
VI | 738 | 1451 | -40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g)....
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