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3 years ago

15

Scoring: Your score will be based on the number of correct matches minus the mumber of incorrect matches. There is no penalty fo

r missing matches Use the References to secess Important values If seeded far this question Predict whether AS for cach reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All I(g) Cl2(g) 21CI(g) PCI5(g PCh(g)+ Cl2(g) AS 0 CO2(g) H2(g)- +H,0(g) CO(g) AS<0 2CO(g)+ 2NO(g) 2CO;(g) N() too close to decide 2H2O2()2H2O(D) O2(g)

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

a. There is little or no change in the entropy.

b. ΔS is greater than zero.

c. There is little or no change in the entropy.

d. ΔS is less than zero.

e. ΔS is greater than zero.

Explanation:

The change in the entropy (ΔS) depends on the change in the amount of gaseous moles, Δn(g) = ngas, products - ngas,reactants.

If Δn(g) > 0, the entropy increases (ΔS > 0)
If Δn(g) < 0, the entropy decreases (ΔS < 0)
If Δn(g) = 0, there is little or no change in the entropy.

<em>a. I₂(g) + Cl₂(g) → 2 ICl(g)</em>

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

<em>b. PCl₅(g) → PCl₃(g) + Cl₂(g)</em>

Δn(g) = 2 - 1 = 1. ΔS is greater than zero.

<em>c. CO₂(g) + H₂(g) → H₂O(g) + CO(g)</em>

Δn(g) = 2 - 2 = 0. There is little or no change in the entropy.

<em>d. 2 CO(g) + 2 NO(g) → 2 CO₂(g) + N₂(g)</em>

Δn(g) = 3 - 4 = -1. ΔS is less than zero.

<em>e. 2 H₂O₂(l) → 2 H₂O(l) + O₂(g)</em>

Δn(g) = 1 - 0 = 1. ΔS is greater than zero.

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The half-reaction for the oxidation of the manganese in $MnCO_3(s)$ to $MnO_2(s)$ .:

$MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+$

Explanation:

$MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;

Step 1: Balance all the atom beside oxygen and hydrogen atom;

$MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

Manganese and carbon are balanced.

Step 2: Balance oxygen atom adding water on the required side:

$MnCO_3+H_2O+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

$MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

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3 years ago

Which one of the following pairs of acids and conjugate bases is incorrectly labeled or incorrectly matched? Acid Conjugate Base

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Answer : The incorrect option is, (d) $NH_4^+/NH_2^-$

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

(a) The equilibrium reaction will be,

$HF+H_2O\rightleftharpoons F^-+H_3O^+$

In this reaction, $HF$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $F^-$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $HF/F^-$ are act as a conjugate acid-base.

(b) The equilibrium reaction will be,

$HClO+H_2O\rightleftharpoons ClO^-+H_3O^+$

In this reaction, $HClO$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $ClO^-$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $HClO/ClO^-$ are act as a conjugate acid-base.

(c) The equilibrium reaction will be,

$H_2O+H_2O\rightleftharpoons OH^-+H_3O^+$

In this reaction, $H_2O$ is an acid that donate a proton or hydrogen to

In this reaction, $H_2O/OH^-$ are act as a conjugate acid-base.

(d) The equilibrium reaction will be,

$NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+$

In this reaction, $NH_4^+$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $NH_3$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $NH_4^+/NH_3$ are act as a conjugate acid-base.

(e) The equilibrium reaction will be,

$H_3O^++H_2O\rightleftharpoons H_2O+H_3O^+$

In this reaction, $H_3O^+$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms

In this reaction, $H_3O^+/H_2O$ are act as a conjugate acid-base.

From this we conclude that that, $NH_4^+/NH_2^-$ pairs of acids and conjugate bases is incorrectly labeled or incorrectly matched.

Hence, the incorrect option is, (d) $NH_4^+/NH_2^-$

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