Class 12th

This is the type of reaction which bheavies like a first order reaction inspite of being a higher order reaction (second or third). This happens due to a particular reactant being present in an excessive quantity thorughout the chemical reaction.

These are the important highlights of integrated rate equations:

• predict the rate of a reaction
• calculate order of reaction (zero, first, second)
• understand mechanism of a reaction
• compute half life
• calculate the value of k

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

 Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13

Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)
y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).
Let's follow the solution:
y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)
Now dy/dx = (1/2√y) * f'(x) = .
The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.
Let's assume y's derivative is taken w.r.t to f(x).
y = -tan?¹x + c
given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2
y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6

f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12

2(cos²θ/sin²θ) - 5/sinθ + 4 = 0
(2sinθ - 1)(sinθ - 2) = 0
sinθ = 1/2 only
∴ θ = π/6, 5π/6
↓↓
θ?, θ?
∫(π/6 to 5π/6) cos²3θdθ = ∫(π/6 to 5π/6) (1+cos6θ)/2 dθ = π/3

6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6

Yes, the phase of the reactant (solid, liquid or gas) actually affects the rate of a reaction. Gases and Liquids tend to move more freely and effectively as compared to solids, which increaes the frequency of particles colliding with each other.

0.50
 y = x²-3x+2, x+y=a, x-y=b
x?=2 x?=1
y? = 4-6+2 = 0 y?=0
(2,0)
(1,0)
b=2
a=1
∴ a/b = 1/2 = 0.5