7–2 Cooling of a Hot Block by Forced Air at High

NO reacts with Br2 in the gas phase according to the following chemical equation: 2NO(g) +Br2(g)2NOBr(g) It is observed that, wh

klemol [59]

Answer:

a) $rate=r=k[NO]^{2} [Br_{2}]^{1}$

b) $k=\frac{1}{s*M^{2}}$

Explanation:

First of all you need to indicate the reaction order of each reactant ( $NO$ and $Br_{2}$ ):

1. $Br_{2}$

Note that if $Br_{2}$ concentration ( $[Br_{2} ]$ ) is reduced to 1/3 of its initial value, the rate of the reaction is also reduced to 1/3 of its initial value, it means:

$[Br_{2} ]=1/3$ then $r=1/3$

As the change in the rate of the reaction is equal to the change of the initial concentration of $Br_{2}$ , you could concluded that the reaction is first order with respect to $Br_{2}$

2. $NO$

Now, note that if $NO$ concentration ( $[NO]$ ) is multiplied by 3.69, the rate of the reaction increases by a factor of 13.6. In this case, to know the ratio could be advisable divide the rate of the reaction (13.6) over the factor whereby was multiplied the concentration (3.69), as follows:

$\frac{13.6}{3.69}=3.69$

As the result is the same factor 3.69 you could concluded that the change of the rate of reaction is proportional to the square of the concentration of A:

$r=[NO]^{2} =3.68^{2} =13.6$

It means that the reaction is second order with respect to $NO$

3. Rate Expression

Remember that the rate expression of the reactions depend on the concentration of each reactant and its order. In this case we have 2 reactants: $NO$ and $Br_{2}$ , then we have a rate law depending of 2 concentrations, as follows:

Note that the expression is the result of the concentration of each reactant raised to its reaction order (previously determined)

<em>Note: I hope that you do not mix up the use of the rates of reaction of each reactant, that is experimentally determined, with the stoichiometric coefficient, are different.</em>

4. Rate constant units (k)

Assuming concentration is expressed as $\frac{mol}{L}=M$ and time is in second, to find the units of k we need to solve an equation with units and with supporting of the rate equation previously obtained, as follows:

$r=k[NO]^{2} [Br_{2}]^{1}$

Where:

$[r]=[\frac{M}{s}]$

$[[NO]]=[M]$

$[ [Br_{2}]]=[M]$

Then:

$\frac{M}{s}=kM^{2} M^{1}$

$\frac{M}{s}=kM^{3}$

$\frac{M}{s*M^{3}}=k$

The units of the rate constant k are:

$k=\frac{1}{s*M^{2}}$

8 0

4 years ago

An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFar

Musya8 [376]

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

6 0

4 years ago

Hãy trình bày sự hiểu biết của bạn về đo dòng điện

Sidana [21]

Answer:

-Khái niệm:

Đo dòng điện là sử dụng các dụng cụ như ôm kê, vôn kế, ampe kế, tần số kế… để xác định các đại lượng vật lý của dòng điện

-Đo lường điện để làm gì?

Phát hiện hư hỏng sự cố trong mạch điện và các thiết bị vi mạch

Xác định các giá trị cần đo

Đánh giá chất lượng của các thiết bị sau sản xuất

Xác định thông số kỹ thuật của thiết bị

-Phân loại dụng cụ đo điện: Hiên nay có 2 phương pháp phân loại chính

a. Theo nguyên lý làm việc

Dụng cụ đo kiểu điện từ

Dụng cụ đo kiểu điện động

Dụng cụ đo kiểu cảm ứng

Dụng cụ đo kiểu từ điện

b. Theo đai lượng, giá trị cần đo

+Đo điện năng: Ví dụ công tơ điện

+Đo điện áp: Ví dụ: Vôn kế

+Đo dòng điện: Ví dụ: Ampe kế

+Đo công suất: Oát kế

+Đo điện trở: Ôm kế

Sai số khi đo: Khi đo lường luôn xảy ra các sai số

Sai số tương đối: Là tỷ lệ % của sự chênh lệch giữa giá trị đo được và giá trị thự

8 0

3 years ago

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operat

Gwar [14]

Answer:

$\eta _{max} = 0.2413 = 24.13%$

$\eta' _{max} = 0.5061 = 50.61%$

Given:

$T_{1max} = 100^{\circ} = 273 + 100 = 373 K$

operating temperature of heat engine, $T_{2} = 10^{\circ} = 273 + 10 = 283 K$

$T_{3max} = 300^{\circ} = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _{max}$ is given by:

$\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}$

$\eta _{max} = 1 - \frac{283}{373} = 0.24$

$\eta _{max} = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_{3max}$ changes to $300^{\circ}$ , so, the new maximum efficiency, $\eta' _{max}$ is given by: