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3 years ago

Use the reaction N2(g) + 3H2(g) 2NH3(g) to answer the following questions

Chemistry

1 answer:

kumpel [21]3 years ago

6 0

Answer:

Explanation:

a ) If N₂(g) and 3H₂(g) is added to the system , 2 moles of additional ammonia will be produced .

b ) If pressure is decreased , less amount of ammonia will be formed, because forward reaction reduces the pressure. So, reaction will take place in reverse direction.

c ) Keq = [ NH₃ ] ² / [ N₂ ] [ H₂]³

d ) Substituting the given values in the equation ,

Keq = [ 6M ] ² / [ 3M] [ 4M]³

= 36 / 3 x 64 M⁻²

= 18.75 x 10⁻² M⁻² .

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Which of the following values would you expect for the ratio of half-lives for a reaction with starting concentrations of 0.05M

harkovskaia [24]

Answer:

The expected ratio of half-lives for a reaction will be 5:1.

Explanation:

Integrated rate law for zero order kinetics is given as:

$k=\frac{1}{t}([A_o]-[A])$

$[A_o]$ = initial concentration

[A]=concentration at time t

k = rate constant

if, $[A]=\frac{1}{2}[A_o]$

$t=t_{\frac{1}{2}}$ , the equation (1) becomes:

$t_{\frac{1}{2}}=\frac{[A_o]}{2k}$

Half life when concentration was 0.05 M= $t_{\frac{1}{2}}$

Half life when concentration was 0.01 M= $t_{\frac{1}{2}}'$

Ratio of half-lives will be:

$\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}$

The expected ratio of half-lives for a reaction will be 5:1.

6 0

3 years ago

Convert 7.89 x 10^8 molecules of water to liters.

NNADVOKAT [17]

Answer:

2.94 x 10⁻¹⁴L

Explanation:

To solve this problem, we have to assume that the condition of this water is at standard temperature and pressure, STP.

At STP;

1 mole of gas have a volume of 22.4L

So, let us find the number of moles of this water first;

6.02 x 10²³ molecules can be found in 1 mole of a substance

7.89 x 10⁸ molecules will contain $\frac{7.89 x 10^{8} }{6.02 x 10^{23} }$ = 1.31 x 10⁻¹⁵mole of water

So;

Volume of water = 22.4 x 1.31 x 10⁻¹⁵ = 2.94 x 10⁻¹⁴L

5 0

3 years ago

Can u help me with #1 #2 #3

nadezda [96]

1.A

2.A i think

3.C

dawdasdwasdwasdwasdwasd

3 0

3 years ago

Read 2 more answers

The mass of an object is the ___.

timama [110]

A) amount of matter it contains.

3 0

3 years ago

Read 2 more answers

Balance the following redox reaction in basic solution. Cl â’ (aq) + MnO â’ 4 (aq) â†’ Cl 2 (g) + MnO 2 (s)

FinnZ [79.3K]

$6\; \text{Cl}^{-} \; (aq) + 2\; \text{MnO}_4^{-} \; (aq) + 4 \; \text{H}_2 \text{O} \; (l) \to 3\; \text{Cl}_2 \; (g) + 2\; \text{MnO}_2 \; (s) + 8\; \text{OH}^{-} \; (aq)$

<h3>Explanation</h3>

The oxidation state of the manganese atom $\text{Mn}$ changes from +7 as in $\text{MnO}_4^{-}$ to +4 as in $\text{MnO}_2$ .
There are one $\text{Mn}$ atom in each $\text{MnO}_4$ ion.
Reducing each $\text{MnO}_4^{-}$ ion would thus consume three electrons.

Similarly, the oxidation state of the chlorine atom $\text{Cl}$ changes from -1 as in $\text{Cl}^{-}$ to 0 as in $\text{Cl}_2$ .
It takes two $\text{Cl}^{-}$ ions to produce one $\text{Cl}_2$ molecule.
Oxidizing every two $\text{Cl}^{-}$ would thus produce one $\text{Cl}_2$ while releasing two electrons.

Three $\text{Cl}_2$ molecules contain six chlorine atoms. Three $\text{Cl}_2$ would thus correspond to six $\text{Cl}^{-}$ ions.

Combining six $\text{Cl}^{-}$ , two $\text{MnO}_4^{-}$ ions, three $\text{Cl}_2$ ions, and two $\text{MnO}_2$ would balance the changes in oxidation state.

$6\; \text{Cl}^{-} \; (aq) + 2\; \text{MnO}_4^{-} \; (aq) \to 3\; \text{Cl}_2 \; (g) + 2\; \text{MnO}_2 \; (s)$ (NOT BALANCED)

Still, the product side lacks four oxygen atoms.

In an acidic environment, oxygen atoms would combine with protons to produce water.
In a basic environment (like this one,) there are nearly no protons for oxygen atoms to combine with. Oxygen atoms will likely combine with water to produce hydroxide ions.

Each oxygen atom combine with a water molecule to produce two hydroxide ions. Adding four water molecules and eight hydroxide ions would balance the equation.

5 0

3 years ago