Question

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ⓐ Prove geometrically $\sf lim \frac{}{x→0} \frac{sinx}{x} = 1$ where x is in radian and hence evaluate $\sf lim \frac{}{x→0} \frac{sin4x}{sin2x}$

ⓑ Differentiate $\sf\frac{ {x}^{5} - cosx}{sinx}$ with respect to x .

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Answer 1

Step-by-step explanation:

Since (A) is the intermediate reactive species whose concentration is determined from equilibrium step.

Slow step is:

A +

B

2

AB+

B

(slow)

r = k[A]

[B

2

]

From equilibrium step

A

2

⇌

A+

A(fast)

k

eq

=

[A

2

]

[A]

2

∴

[A]=

(k

eq

[A

2

]

1/2

Substitute the value of [A] in equation (i),

r=

k.k

eq

1/2 [A

2

]

1/2

[B

2

]

Thus, order of reaction =

2

1

+

1=

1

2

1