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arsen [322]
2 years ago
8

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Chemistry
1 answer:
Karo-lina-s [1.5K]2 years ago
5 0

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

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To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

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