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2 years ago

For the reaction: N2(g) + H2(g) → NH3(g)

Chemistry

1 answer:

gayaneshka [121]2 years ago

3 0

The rate of increase of NH3 is 0.22M/s.

<h3>What is the balanced equation of the reaction?</h3>

The balanced equation of the reaction is given below:

N2(g) + 3 H2(g) → 2 NH3(g)

The rate of decrease of N2 is half the rate of increase of NH3.

Rate of decrease of N2 = -0.11 M/s

Rate of increase of NH3 = 2 × +0.11 M/s = 0.22M/s.

In conclusion, the rate of formation of products is dependent on the rate of disappearance of reactants.

Learn more about rate of reaction at: brainly.com/question/25724512

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The answer to your question is letter D.

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A. Has a fixed volume This is not the right answer, liquids and gases take the shape of the container in which they are.

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Which statement best describes organs in an organ system? A. Each organ does part of a larger job. B. Each organ can perform onl

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The correct answer is - A. Each organ does part of a larger job.

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A group of organs makes an organ system to perform a particular but large function in the organism for its survival. An example of the organ in an organ system is the heart in the cardiovascular system. The heart is an organ that pumps the blood out of the heart to the various part of the cardiovascular system such as lungs, arteries, and veins so it can take nutrients and oxygen to various parts carried by the blood.

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3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

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