Ask question

Ask question

All categories

1 year ago

11

Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2

.5 x 10-1 M, calculate the reaction quotient (Q) and determine which way the chemical system would shift by comparing the value of Q to K.

1 answer:

Brilliant_brown [7]1 year ago

8 0

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

$K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}$

And Q, is:

$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

$Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}$

<h3>Q = 1.6x10⁻⁷</h3>

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

brainly.com/question/24301138

You might be interested in

If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev

shepuryov [24]

Answer:

It can be removed by acidic chemicals

Explanation:

8 0

2 years ago

Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect,

SVEN [57.7K]

Answer: 0.0043mole

Explanation:Please see attachment for explanation

3 0

2 years ago

How many moles of lead, Pb, are in 1.50 x 1012 atoms of lead?

SpyIntel [72]

Answer:

2.49*10⁻¹² mol

Explanation:

Use Avogadro's Number for this equation (6.022*10²³). Divide Avogrado's by the number of atoms you have to find moles. You are answer should be 2.49*10⁻¹² mol.

7 0

1 year ago

Read 2 more answers

COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at

mestny [16]

<u>Answer:</u> The value of $K_p$ for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of $COCl_2$ = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial: 1 - -

At eqllm: 1-x x x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

$[(1 - x) + x+ x]=1.2\\\\x=0.2atm$

The expression for $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

$p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm$

Putting values in above equation, we get:

$K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05$

Hence, the value of $K_p$ for the reaction at 690 K is 0.05

3 0

1 year ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

1 year ago