Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
Answer:
Any 4 characteristic of catalyst are:
1. Catalysts remain unchanged at the end of chemical reaction in their mass and structure.
2. They donot start the chemical reaction.
3. They change the rate of the chemical reaction.
4. Catalyst is specific to the chemical reaction. It means that one catalyst can alter the rate of only one chemical reaction.
Answer:
First, balance the half-reactions
Second, equalize the electrons
Third,add two reaction equations to get final answer
Explanation:
For example
H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺
(i) Balancing the half reactions
H₂C₂O₄-------->2CO₂+2H⁺+2e⁻
5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O
(ii)
Equalizing the electrons
5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5
10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2
(iii)
Add the two where electrons and some Hydrogen ions will cancel out
5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O
Chemical Reactions and Moles of Reactants and Products
That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide.
