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2 years ago

2A + B + 2C ? D + E

Chemistry

1 answer:

Amanda [17]2 years ago

8 0

Answer:

C Rate $=k [A]^{2} [B][C]^{-1}$

Explanation:

In order to determine the correct rate law, let's use Trial 1 as baseline. Therefore:

An increase in [A] in Trial 2 by a factor of $\sqrt{2}$ leads to an increase in the rate of reaction by a factor of 2 (i.e. the reaction rate is doubled). Thus, there is second order in [A].

Similarly,

An increase in [B] in Trial 3 by a factor of 1.667 leads to an increase in the rate of reaction by a factor of 1.667. Thus, there is first order in [B].

Futhermore,

An increase in [C] in Trial 4 by a factor of 1.71 leads to a decrease in the rate of reaction by 1.71. Thus, there is inverse first order in [C].

Therefore, the correct rate law is:

Rate $=k [A]^{2} [B][C]^{-1}$

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The average kinetic energy and rms speed of N₂ molecules at STP is $5.65686 \times 10^{-21}$ and $493 \mathrm{~m} / \mathrm{s}$

Given,

$\begin{aligned}&\mathrm{P}=1.013 \times 10^{5} \mathrm{~Pa} \\&\mathrm{~T}=273.15 \mathrm{~K} \\&\rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}\end{aligned}$

The average kinetic energy of a molecule is given by, $K . E .=\frac{3}{2} k T$ where k is the Boltzmann constant and Tis the absolute temperature of the gas.

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$K.E=$5.65686 \times10^{-21}$$

The rms speed of $\mathrm{N}_{2}$ molecules is given by

$\begin{aligned}&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.25}}=493 \mathrm{~m} / \mathrm{s}\end{aligned}$

The average kinetic energy of a gas's particles is inversely related to its temperature. As the gas warms, the particles must travel more quickly since their mass is constant.

The average kinetic energy (K) is equal to one half of the mass (m) of each gas molecule times the RMS speed (vrms) squared.

Learn more about average kinetic energy brainly.com/question/1599923

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