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erica [24]
3 years ago
8

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

es only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a tank with of ammonia gas and of oxygen gas, and when the mixture has come to equilibrium measures the amount of water vapor to be . Calculate the concentration equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to significant digits.
Chemistry
1 answer:
Gala2k [10]3 years ago
5 0

\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

\text{and in the liquid form it is easily transported. An industrial chemist studying this}

\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

\text{to be} \  \mathbf{2.6\  mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}

\text{ammonia at the final temperature of the mixture. Round your answer to  2 significant digits.}

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

<u>Answer:</u>

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

<u>Explanation:</u>

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is ns^{1-2}

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is np^{1-6}

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is [(n-1)d^{1-10}ns^{0-2}]

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is [(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]. They are also known as lanthanide and actinide series.

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As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 2:</u> Am

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As, the last electron is entering the (f subshell), it is a inner transition element.

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Indium is the 49th element of the periodic table having electronic configuration of [Kr]5s^25p^1

As, the last electron is entering the p subshell, it is a main group element.

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As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

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As, the last electron is entering the p subshell, it is a main group element.

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Answer:

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