Chemistry, 21.07.2019 18:30 nsjcbdhcb

Δs is positive for the reaction a.2no (g) + o2 (g) → 2no2 (g) b.2n2 (g) + 3h2 (g) → 2nh3 (g) c. c3h8 (g) + 5 o2 (g) → 3co2 (g) + 4 h2o (g) d. mg (s) + cl2 (g) → mgcl2 (s) e. c2h4 (g) + h2 (g) → c2h6 (g)

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Answer from: gonzalesalexiaouv1bg

Option (c) is the correct answer.

Explanation:

Entropy means the degree of randomness present in a substance or within the reactants in a chemical reaction.

Change in entropy is represented by $\Delta S$ . More is the degree of randomness present more positive will be the value of $\Delta S$ . Similarly, less is the degree of randomness present within a substance lesser will be the value of $\Delta S$ .

(a) $2NO(g) + O_{2}(g) \rightarrow 2NO_{2}(g)$

Here, 3 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

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

Here, 5 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

Here, 6 moles of reactants are giving 7 moles of product. Hence, entropy is increasing so, the value of $\Delta S$ is positive.

(d) $Mg(s) + Cl_{2}(g) \rightarrow MgCl_{2}(s)$

Here, 2 moles of reactants are giving 1 mole of product. Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

(e) $C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)$

Here, 2 moles of reactants are giving 1 mole of product. Hence, entropy is decreasing so, the value of $\Delta S$ is negative.

Thus, we can conclude that $\Delta S$ is positive for the reaction $C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)$ .

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

answer: e. sour taste

explanation: acids are those substances which dissociates in water to give $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

$hcl\rightarrow h^++cl^-$

bases are those substances which dissociates in water to give $oh^-$ ions or accepts $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

$naoh\rightarrow na^++oh^-$

$nh_3+h^+\rightarrow nh_4^+$

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Δs is positive for the reaction a.2no (g) + o2 (g) → 2no2 (g) b.2n2 (g) + 3h2 (g) → 2nh3 (g) c. c3h...

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