Class 11th
Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
8 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
8 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
8 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
8 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
8 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
8 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
8 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.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 684k Reviews
- 1800k Answers
