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New answer posted
11 months agoContributor-Level 10
An acid-base pair which differs by a proton only is known as conjugate acid-base pair. For example, HCl, and Cl represents conjugate acid base pair.
The conjugate acid/base for the species HNO2, CN–, HClO4F–, OH–, CO32-, and S2- are NO2–, HCN, ClO4–, HF, H2O, HCO3– and HS–.respectively.
New answer posted
11 months agoContributor-Level 10
Kc = [CH4] [H2O] / [CO] [H2]3
3.90 = [CH4]*0.02 / (0.30) (0.10)3
[CH4] = 3.90*0.30* (0.10)3 / 0.02
=5.85*102 M
Thus, the concentration of methane in the mixture is 5.85*102 M.
New answer posted
11 months agoContributor-Level 10
The equilibrium constant expression is
Kc = [O3]2 / [O2]3
2.0*1050 = [O3]2 / (1.6*102)3
[O3]2=2.0*1050* (1.6*102)3
=8.192*1056
Hence, the equilibrium concentration of ozone is
[O3]= 2.86*1028M
New answer posted
11 months agoContributor-Level 10
Following conclusions can be drawn from the values of Kc .
(a) Since the value of Kc is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of Kc is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of Kc is 1.8, this means that both the products and reactants have appreciable concentration.
New answer posted
11 months agoContributor-Level 10
Let the partial pressures of CO and H2 be p.
CO (g) + H2?O(g) ? CO2?(g) + H2?(g)
Initial conc. 4 bar 4 bar 0 0
At equilibrium 4-p 4-p p
New answer posted
11 months agoContributor-Level 10
(a) The equilibrium constant expression is
Kc = [PCl3] [Cl2] / [PCl5]
(b) The equilibrium constant expression for the reverse reaction is
Kc (reverse) = 1/ (8.3*103) =120.48
(c) (i) By adding more of PCl5, value of Kc will remain constant because there is no change in temperature.
(ii) When pressure is increased, the value of the equilibrium constant remains unaffected.
(iii) By increasing the temperature, the forward reaction will be favoured since it is endothermic in nature. Therefore, the value of equilibrium constant will increase.
New answer posted
11 months agoContributor-Level 10
(i) Equilibrium will be shifted in the forward direction.
(ii) Equilibrium will be shifted in the backward direction.
(iii) Equilibrium will be shifted in the backward direction.
(iv) Equilibrium will be shifted in the forward direction.
New answer posted
11 months agoContributor-Level 10
(a) The expression for Kp for the reaction is:
Kp = (pCO) x (pH2) / (pCH4) x (pH2O)
(b) (i) By increasing the pressure, the number of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of Kp will decrease.
(ii) If the temperature is increased, according to Le Chatelier's principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of Kp will increase.
(iii) The addition of catalyst will not change t
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