(a) Moles of H3O+ = \(\frac{25 × 0.1}{1000}\) = .0025 mol
Moles of OH- = \(\frac{10 × 0.2 × 2}{1000}\) = .0040 mol
Thus, an excess of OH- = .0015 mol
[OH-] =\(\frac{ .0015}{35 × 10^{-3}}\) mol/L = .0428
pOH = -log[OH] = 1.36
pH = 14 - 1.36 = 12.63 (not matched)
(b) Moles of H3O+ =\(\frac{10 × 0.01 × 2}{1000}\) = .0002 mol
Moles of OH- = \(\frac{10 × 0.1 × 2}{1000}\) = .0002 mol
Since there is neither an excess of H3O+ or OH-, the solution is neutral. Hence, pH = 7.
(c) Moles of H3O+ = \(\frac{10 × 0.1 × 2}{1000}\)= .002 mol
Moles of OH- = \(\frac{10 × 0.1 }{1000}\) = 0.001 mol
Excess of H3O+ = .001 mol
Thus, [H3O+] = \(\frac{.001}{20 × 10^{-3}}\) = \(\frac{10^{-3}}{20 × 10{-3}}\) = 0.5
∴ pH = -log(0.05) = 1.30
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
Medium Kc/Kp values indicate optimum product formation.
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.