The activation energy Ea for a particular reaction is 42.2 kJ/mol. How much faster is the reaction at 343 K than at 321 K? (R = 8.314 J/mol • K)

Answer 1

You can solve this by eventually solving for v^2 and then v and then comparing

You can use dimensional analysis and if all of the same substances are in the reaction then you can apply them across the board, include activation energy and the equation:

KE=1/2m*v^2

Use dimensional analysis and mind your units:

For temperature of 343K:

343K* 8.314J/(mol*K) Kelvin cancel top to bottom and you have:

2851.702J/mol

For 321K:

321K* 8.314J/(mol*K) =

2668.794J/mol

Now take these to the activation energy. Mind your units because you have Joules and you need kJ or kilojoules.

1kJ=1000J

42.2kJ/mol*(1000J/1kJ)=

42200J/mol

Now combine energies so you have the same units and then Multiply by 1 mol for each so you just have Joules.

42200J/mol + 2851.702J/mol=

45051.702 J/mol*1mol

42200J/mol + 2668.794J/mol= 44868.794J/mol *1 mol

Now multiply by 1mol each to get just Joules, one of the energy units (the other one is Hz or Hertz) but the units Joules applies here.

Now evaluate

KE=1/2mv^2

using 1gram for both. But kilograms for the sake of this equation:. 1g/1000 = 0.001kg

45051.702= 1/2(0.001)v^2

45051.702= 0.0005v^2

90103404= v^2

9492.48m/s= v

44868.794= 0.0005v^2

89737588= v^2

9472.99m/s= v

Now take the difference:

9492.48- 9472.99 = 19.49m/s faster.

And you know that your units are correct because:

1Joule= 1N*m= 1Kg*m^2*s^-2