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gizmo_the_mogwai [7]
3 years ago
10

A Questic

Chemistry
1 answer:
Ivanshal [37]3 years ago
3 0

Answer:

No, it is not feasible because the Gibbs free energy change is positive

Explanation:

∆Hreaction= (-602 KJ/mol) - (-348 KJ/mol) = -254 KJ/mol

∆Sreaction = (42 + 27) J/Kmol - (33 + 44) J/K = -8J/Kmol

From;

∆G = ∆H - T∆S

∆G = 254 × 10^3 J/mol - [340K × (-8 J/Kmol)]

∆G = 2.57 × 10^5 J/mol

Note that when the change in free energy is positive, a reaction is non spontaneous. Only a reaction that has a negative change in free energy is spontaneous.

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A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K
charle [14.2K]

<u>Answer:</u> The equilibrium constant for the total reaction is 4.09\times 10^{-6}

<u>Explanation:</u>

We are given:

K_{c_1}=0.282\\\\K_{c_2}=41

We are given two intermediate equations:

<u>Equation 1:</u> N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282

The expression of K_{c_1} for the above equation is:

K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}

0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}        .......(1)

<u>Equation 2:</u> H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41

The expression of K_{c_2} for the above equation is:

K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}

41=\frac{[HI]^2}{[H_2][I_2]}       ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}

Now, dividing expression 1 by expression 2, we get:

\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}

\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}

The above expression is the expression for equilibrium constant of the total equation, which is:

2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c

Hence, the equilibrium constant for the total reaction is 4.09\times 10^{-6}

8 0
3 years ago
which of these pieces of equipment would be the most appropriate for precisely measuring 29 mL of liquid? Explain your reasoning
Anna007 [38]

Answer:

The best equipment would be the graduated cylinder. Why?

Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.

Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.

Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.

Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.

We are left with the graduated cylinder, and that is our answer.

Explanation:

Hope this helped!

5 0
3 years ago
Which of the following conclusions might a scientist make when fossils of two different species of very similar organisms are fo
nata0808 [166]

Answer:

if its multiple, C and D, if not, C

Explanation:

brainliesttt?

5 0
3 years ago
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