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3 years ago

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

Chemistry

1 answer:

harina [27]3 years ago

6 0

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

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A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$

6 0

3 years ago

Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when di

Gwar [14]

Answer:

NaCl neutral

Na₂SO₄ basic

NaHCO₃ basic

K₃PO₄ basic

Na₃C₆H₅O₇ basic

CaCl₂ neutral

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water.

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water

Explanation:

<em>Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs.</em>

<em />

NaCl

When NaCl is dissolved in water the resulting solution is neutral.

Na₂SO₄

When Na₂SO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the SO₄²⁻ ion.

SO₄²⁻(aq) + H₂O(l) ⇄ HSO₄⁻(aq) + OH⁻(aq)

NaHCO₃

When NaHCO₃ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the HCO₃⁻ ion.

HCO₃⁻(aq) + H₂O(l) ⇄ H₂CO₃(aq) + OH⁻(aq)

K₃PO₄

When K₃PO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the PO₄³⁻ ion.

PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

Na₃C₆H₅O₇ (sodium citrate)

When Na₃C₆H₅O₇ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the C₆H₅O₇³⁻ ion.

C₆H₅O₇³⁻(aq) + H₂O(l) ⇄ C₆H₆O₇²⁻(aq) + OH⁻(aq)

CaCl₂

When CaCl₂ is dissolved in water the resulting solution is neutral.

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water according to the following equation:

Na₂C₂O₄(aq) ⇄ 2 Na⁺(aq) + C₂O₄²⁻(aq)

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water according to the following equation:

C₂H₂O₄(aq) ⇄ C₂HO₄⁻(aq) + H⁺(aq)

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Answer:

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2 years ago

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Answer:

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Explanation:

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3 years ago