2 years ago

What does a ka value of approximately 1 show about the equilibrium?

Chemistry

1 answer:

Artemon [7]2 years ago

8 0

I think d I’m not sure for sure

You might be interested in

Need help with chemistry hw please help

Andru [333]

Co2 = two covalent bonds
ccl4 = 4 covalent bonds
Lih = covalent bond

5 0

2 years ago

Read 2 more answers

Does this sound good/ right? I hope so! Thank you! please give me criticism NICELY!!

Mamont248 [21]

This sounds good so far.
May want to check how you say "and are spread way out" - a better way to say it would be "the crystals are spread out"

5 0

2 years ago

Read 2 more answers

A gas that produces a "pop" with a burning splint is carbon dioxide. true or false

Ksivusya [100]

False, a gas that produces a pop is hydrogen
Hope this helped :)

7 0

2 years ago

True or False: As the ball rises from point 1 to point 3, it slows down

quester [9]

True, it would slow down.

7 0

2 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.

patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of $NH_{3}$ is as follows:.

$2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)$

And, initially only $NH_{3}$ is present.

The given data is as follows.

$P_{NH_{3}}$ = 2.3 atm at equilibrium

$P_{H_{2}} = 3 \times P_{N_{2}}$ = 0.69 atm

Therefore,

$P_{N_{2}} = \frac{0.69 atm}{3}$

= 0.23 aatm

So, $P_{NH_{3}}$ = 2.3 - 2(0.23)

= 1.84 atm

Now, expression for $K_{p}$ will be as follows.

$K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}$

$K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}$

= $\frac{0.23 \times 0.33}{3.39}$

= 0.0224

or, $K_{p} = 2.2 \times 10^{-2}$

Thus, we can conclude that the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is $2.2 \times 10^{-2}$ .

6 0

3 years ago