The unit of the rate constant in a 1st Order reaction rate equation (NOT the 'Arrhenius equation', as stated in the question) is One over Time.
General form of a reaction rate equation :
rate (mol.L-1.time-1) = [rate constant(Ln-1.mol1-n.time-1)]*[Concentration()]n
where:
* n is the Order of the rate equation (that is of the rate limiting step) * all units are (italicalised) between brackets It can easily be seen in this that for n=1 (1st Order) the equation is:
r = k * C1
and in units:
mol.L-1.time-1 = (L0.mol0.time-1)*(mol.L-1)1
so:
(mol.L-1.time-1) = (time-1)*(mol.L-1)
Only the value of the rate constant k is depending on temperature only (cf. Arrhenius equation), though temperature is NOT in its unit.
This question can only be answered if 'Temperature' is read as 'Time', because temperature is not in the rate (constant) of a reaction as 1/Temp. For this one should review the Arrhenius equation itself:
k=Ae-[Ea/RT]
The unit of the rate constant in a 1st Order reaction rate equation (NOT the 'Arrhenius equation', as stated in the question) is One over Time.
General form of a reaction rate equation :
rate (mol.L-1.time-1) = [rate constant(Ln-1.mol1-n.time-1)]*[Concentration()]n
where:
It can easily be seen in this that for n=1 (1st Order) the equation is:
r = k * C1
and in units:
mol.L-1.time-1 = (L0.mol0.time-1)*(mol.L-1)1
so:
(mol.L-1.time-1) = (time-1)*(mol.L-1)
Only the value of the rate constant k is depending on temperature only (cf. Arrhenius equation), though temperature is NOT in its unit
Temperature and activation energy
Temperature and activation energy - apex
You need to use the Arrhenius equation to solve this kind of problem. Since you haven't given the activation energy, we can't answer it for you.
Pressure x Volume = Constant (at a constant temperature).
I'd use a graph showing an exponential decrease: as pressure increases, volume decreases.
Temperature and activation energy
Temperature and activation energy - apex
Yes, since the R constant has units of Latm/molK, temperature must be in K
If it is in a y=mx+b format. Also, if there is a slope and a constant in the equation.
The equation is xy = k where k is the constant of variation. It can also be expressed y = k over x where k is the constant of variation.
The equation is pV=k (k is a constant at constant temperature).
It is an equation that relates the speed at which a chemical reaction progresses with the activation energy and the temperature of the reactants and products. k = A * e^(-Ea/(R*T)) Where k = velocity constant (different for each reaction) A = pre-exponential factor Ea = activation energy R = universal gas constant (=8,314J/molK) T = temperature
xY = 6.
Temperature and activation energy - apex
A linear non-proportional relationship can be identified from a table if the ratios of the y-values to the x-values are not constant. In other words, if the values in the y-column do not increase or decrease by the same factor for each increase in the x-values. From a graph, a non-proportional linear relationship can be identified if the line does not pass through the origin (0,0) or if the slope of the line is not constant. Finally, in an equation, a non-proportional linear relationship can be identified if it does not have a multiplier or constant ratio in front of the x-variable.
You need to use the Arrhenius equation to solve this kind of problem. Since you haven't given the activation energy, we can't answer it for you.
If the equation of a hyperbola is ( x² / a² ) - ( y² / b² ) = 1, then the joint of equation of its Asymptotes is ( x² / a² ) - ( y² / b² ) = 0. Note that these two equations differ only in the constant term. ____________________________________________ Happy To Help ! ____________________________________________