The question is incomplete. The complete question is :
A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)
Solutions :
If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.
Let's take :


The rate constant =
respectively.
The activation energy and the Arhenius factor is same.
So by the arhenius equation,
and 




Given,
J/mol
R = 8.314 J/mol/K





∴ 
So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.
STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P
V2/V1= T2*P1/T1*P2
V2/V1=273.15K* 90^10^3Pa/ 300.15K * 1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml
Answer:
Biochemical compounds make up the cells and other structure of organism and carry out life processes.Carbon is the basic of all biochemical compounds so carbon carried life process on the earth
<u>0.12 atm</u><u> </u><u>vapor pressure</u><u> of ethanol at 45.0 C.</u>
What is vapor pressure in science definition?
- Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
- The temperature at which the vapour pressure at the surface of a liquid becomes equal to the pressure exerted by the surroundings is called the boiling point of the liquid.
We will use the Clausius-Clapeyron equation,
ln(P2/P1) = dHvap/R[1/T1-1/T2]
where,
P1 = unknown
P2 = 1 atm
T1 = 30 oC = 30 + 273 = 303 K
T2 = 78.3 oC = 78.3 + 273 = 351.3 K
dHvap = 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/K.mol
Feed values,
ln(1/P1) = 39300/8.314[1/303 - 1/351.3]
P1 = 0.12 atm
thus, the vapor pressure at 30° C is 0.12 atm.
Learn more about vapor pressure
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