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melamori03 [73]
4 years ago
15

The formation of nitric oxide (N2(g) + O2(g) Imported Asset 2NO(g)) has a Kc value of 4.7 × 10-31. A classmate claims that this

reaction must really make a lot of nitric oxide since the exponent is such a large value. Explain to the student why her reasoning is correct or flawed.
Chemistry
1 answer:
riadik2000 [5.3K]4 years ago
5 0
Answer is: her reasoning is flawed, because <span>Kc is very small, so the concentration of nitric(II) oxide is also very small. </span>

Balanced chemical reaction: N₂(g) + O₂(g) ⇄<span> 2NO(g).
</span>The equilibrium constant<span> (Kc) is a ratio of the concentration of the products (in this reaction nitrogen(II) oxide) to the concentration of the reactants (in this reaction nitrogen and oxygen):
</span>Kc = [NO]² / [N₂] · [O₂].
Kc = 4.7·10⁻³¹.
If we take equilibrium concentration of oxygen and nitrogen to be 1 M:
[N₂] = [O₂] = 1 M.
[NO] = √[N₂] · [O₂] · Kc.
[NO] = 6.855·10⁻¹⁶ M; equilibrium concentration of nitric oxide.
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Simora [160]

Answer:

Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)

Explanation:

Let's consider the following molecular equation.

Mg(s) + SnSO₄(aq) ⇄ MgSO₄(aq) + Sn(s)

The full ionic equation includes al the ions and the species that do not dissociate in water.

Mg(s) + Sn²⁺(aq) + SO₄²⁻(aq) ⇄ Mg²⁺(aq) + SO₄²⁻(aq) + Sn(s)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the species that do not dissociate in water.

Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)

3 0
4 years ago
9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
WITCHER [35]

Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

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