Divide both sides by (ln(1.00x10^{15}) - ln(1.0x10^{14}) )

3,000/ R / (ln(1.00x10^{15}) - ln(1.0x10^{14}) ) = T

Using R = 8.31 J/mol K , I got a value of 156.7K for T

R is the Universal Gas Constant which is known, so the only thing in the above equation that is not know is T. Solve the equation for T.

Question 3.

Increasing the pressure (or concentration) by a factor of 10 increased the rate by a factor of 10 as well. This indicates that the reaction is first order

We are looking for a single temperature, so T must be the same in both equations.

Also k1 = k2, therefore

(1.0 x10

^{14})exp(=-55,000/RT) = (1.0 x 10^{15}) exp(-58,000/RT)or using another form of the equation:

Settting ln(k1) = ln (k2) yields

-55,000/R (1/T) + ln (1.00x10

^{14}) = -58,000/R (1/T) + ln(1.0 x 10^{15})Bringing the 1/T factors to the left side and the ln values to the right side gives

3,000/R (1/T) = ln(1.00x10

^{15}) - ln(1.0x10^{14})Multiply both sides by T yields

3,000/R = (ln(1.00x10

^{15}) - ln(1.0x10^{14})) TDivide both sides by (ln(1.00x10

^{15}) - ln(1.0x10^{14}) )3,000/ R / (ln(1.00x10

^{15}) - ln(1.0x10^{14}) ) = TUsing R = 8.31 J/mol K , I got a value of 156.7K for T

R is the Universal Gas Constant which is known, so the only thing in the above equation that is not know is T. Solve the equation for T.

Question 3.

Increasing the pressure (or concentration) by a factor of 10 increased the rate by a factor of 10 as well. This indicates that the reaction is first order