# The rate constant of a first order reaction is #4.60 xx 10^(-4) "s"^(-1)# at #350^@ "C"#. If the activation energy is #"104 kJ/mol"#, what is the temperature at which the rate constant is #8.80 xx 10^(-4) "s"^(-1)#?

##### 1 Answer

I get

Let's look at what we have:

#k_1 = 4.60xx10^(-4) "s"^(-1)# at#T_1 = 350^@ "C"# , the rate constant for the reaction at temperature#1#

#k_2 = 8.80xx10^(-4) "s"^(-1)# at#T_2 = ?# , the rate constant for thesamereaction at temperature#2#

#E_a = "104 kJ/mol"#

**THE ARRHENIUS EQUATION RELATES RATE CONSTANTS TO TEMPERATURE AND THE ACTIVATION ENERGY**

These are related via the **Arrhenius equation**, which, for a given rate constant

#\mathbf(k_i = Ae^(-E_a"/"RT_i))# where:

#A# is thepre-exponential factor, and it is basically an experimentally-acquired constant correlating with collision frequency.#R# is theuniversal gas constant. With energy units in#"J"# , we have#R = "8.314472 J/mol"cdot"K"# .#T# istemperaturein#"K"# .#E_a# , as you can tell, is theactivation energy, or the energy needed to help the reactants make it over the transition state "hill" and promote a successful reaction.

To determine the temperature associated with ** cancelling out** the pre-exponential factor (given that for the same reaction, despite being at different temperatures,

#[k_2 = cancel(A)e^(-E_a"/"RT_2)]/[k_1 = cancel(A)e^(-E_a"/"RT_1)]#

Also note that for the same reaction, **the activation energy does not change with temperature**. Instead, the *average number of particles* in an ensemble that have *enough* energy is **higher**.

**DETERMINING THE Y = MX + B FORM OF THE ARRHENIUS EQUATION**

We proceed to derive what would be the graphable form of this equation.

#(k_2)/(k_1) = (e^(-E_a"/"RT_2))/(e^(-E_a"/"RT_1))#

Taking the natural logarithm allows a separation of terms, and a cancellation of exponentials via the fact that

#ln\frac(k_2)(k_1) = ln\frac(e^(-E_a"/"RT_2))(e^(-E_a"/"RT_1))#

#= ln(e^(-E_a"/"RT_2)) - ln(e^(-E_a"/"RT_1))#

#= -(E_a)/(RT_2) - (-(E_a)/(RT_1))#

#= -(E_a)/(RT_2) + (E_a)/(RT_1)#

#= (E_a)/R[-1/(T_2) + 1/(T_1)]#

So that gives us:

#color(blue)(ln\frac(k_2)(k_1) = -(E_a)/R[1/(T_2) - 1/(T_1)])#

*This is the graphable form of the Arrhenius equation, and would normally be used with known temperatures to determine the activation energy.*

**DETERMINING THE NEW TEMPERATURE**

However, we have been solving for

#T_2 = 1/[ 1/(T_1) - (Rln\frac(k_2)(k_1))/(E_a)]#

Yikes, kind of messy. So let's simplify that some more... Cross-multiply:

#T_2 = 1/[ (E_a)/(E_aT_1) - (T_1Rln\frac(k_2)(k_1))/(E_aT_1)]#

Combine in the denominator:

#T_2 = 1/[(E_a - T_1Rln\frac(k_2)(k_1))/(E_aT_1)]#

Now flip! Much nicer.

#color(blue)(T_2 = (E_aT_1)/(E_a - T_1Rln\frac(k_2)(k_1)))#

So, finally, let's get

#color(blue)(T_2) = ["104000 J/mol"cdot("350+273.15 K")]/("104000 J/mol" - ("350+273.15 K")cdot"8.314472 J/mol"cdot"K"cdotln\frac(8.80xx10^(-4) "s"^(-1))(4.60xx10^(-4) "s"^(-1)) )#

#= (64807600 cancel("J/mol")cdot"K")/([104000-(5181.1632268*0.648695)] cancel("J/mol"))#

#=# #"643.96 K"#

#-> color(blue)(~~ 371^@ "C")#

So, this is saying that a **higher** temperature **increases** the rate constant, which makes sense.

The **higher** the rate constant, the **faster** the reaction occurs (given that it is in reciprocal time units, where less time is faster), and the reaction should be **faster** because the average kinetic energy of the system is **higher**, promoting **more** successful collisions **more** quickly.

**CHECKING OUR ANSWER**

And to make sure we didn't botch our calculations, let's check to see we get the right

#ln\frac(k_2)(k_1) = -(E_a)/R[1/(T_2) - 1/(T_1)]#

#k_2 = k_1e^(-(E_a)/R[1/(T_2) - 1/(T_1)])#

So, we have:

#color(green)(k_2) = 4.60xx10^(-4) "s"^(-1)cdote^(-"104000 J/mol"/("8.314472 J/mol"cdot"K")(1/("643.96 K") - 1/("623.15 K"))#

#= 8.7997xx10^(-4) "s"^(-1) ~~ color(green)(8.80xx10^(-4) "s"^(-1))#

It works!