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^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d) If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Explanation:
Sucrose + 
fructose+ glucose
The rate law of the reaction is given as:
![R=k[H^+][sucrose]](https://tex.z-dn.net/?f=R%3Dk%5BH%5E%2B%5D%5Bsucrose%5D)
![[H^+]=0.01M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01M)
[sucrose]= 1.0 M
![R=k[0.01M][1.0 M]](https://tex.z-dn.net/?f=R%3Dk%5B0.01M%5D%5B1.0%20M%5D)
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
![R'=[0.01 M][2.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B2.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B2.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
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]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B0.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B0.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
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 is changed to 0.001 M = R'
![R'=[0.0001 M][1.0 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.0001%20M%5D%5B1.0%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.0001%20M%5D%5B1.0M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration of 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 both are changed to 0.1 M = R'
![R'=[0.1M][0.1M]](https://tex.z-dn.net/?f=R%27%3D%5B0.1M%5D%5B0.1M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.1M%5D%5B0.1M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
2-butane is the correct answer
The reaction involved in this problem is called the combustion reaction where a hydrocarbon reacts with oxygen to product carbon dioxide and water. The reaction of C2H5OH would be as follows:
C2H5OH + 3O2 = 2CO2 + 3H2O
To determine the number of molecules of CO2 that is formed, we need to determine the number of moles produced from the initial amount of C2H5OH and the relation from the reaction. Then we multiply avogadros number which is equal to 6.022x10^23 molecules per mole.
2.00 g C2H5OH ( 1 mol C2H5OH / 46.08 g C2H5OH ) ( 2 mol CO2 / 1 mol C2H5OH ) = 0.0868 mol CO2
0.0868 mol CO2 ( 6.022x10^23 molecules / mol ) = 5.23x10^22 molecules CO2
