Chemistry, 26.07.2019 03:20, rbull7072

In the first step of the ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temperature reaction 4 nh31g2 + 5 o21g2 ¡ 4 no1g2 + 6 h2o1g2 (a) how is the rate of consumption of o2 related to the rate of consumption of nh3? (b) how are the rates of formation of no and h2o related to the rate of consumption of nh3

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answered: hazefreedom3885

Solution: The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

If we have a hypothetical equation:

$A+2B\rightarrow 3C+5D$

Then the rate could be written as:

$rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}$

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

$rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}$

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of O_2 related to rate of consumption of NH_3 as:

$-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

multiply both sides by -5

$\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, rate of consumption of oxygen is $\frac{5}{4}$ the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

$\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 4

$\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}$

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of H_2O to the rate of consumption of ammonia would be:

$\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 6

$\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, the rate of formation of H_2O is $\frac{6}{4}$ times that is 1.5 times to the rate of consumption of ammonia.

answered: Guest

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

$^{m}_{4}x$ and $^{m}_4y$

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

$^{m}_{5}x^{-}$

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be $^{9}_{5}be^{-}$ and $^{9}_{4}be$ .

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.

answered: Guest

answer is: naturally occurring solid, inorganic compound or element is mineral.

minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

for example magnetite is a mineral of iron oxide (fe₂o₃).

calcite is a carbonate mineral and the most stable polymorph of calcium carbonate (caco₃).

dolomite is an carbonate mineral composed of calcium magnesium carbonate (camg(co₃)₂).