The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

how many grams of ammonium carbonate are needed to decompose in order to produce 6.52 g of carbon dioxide?

defon

Answer:

= 14.24g of $(NH4)_{2}CO_{3}$ is required.

Explanation:

Reaction equation:

$(NH4)_{2}CO_{3}$ → $2NH_{3} + CO_{2} + H_{2} O$

Mole ratio of ammonium carbonate to carbon dioxide is 1:1

1 mole of CO2 - 44g

?? mole of CO2 - 6.52g

= 6.52/44 = 0.148 moles was produced from this experiment.

Therefore, if 1 mole of $(NH4)_{2}CO_{3}$ - 96.09 g

0.148 mol of $(NH4)_{2}CO_{3}$ -- ?? g

=0.148 × 96.09

= 14.24g of $(NH4)_{2}CO_{3}$ is required.

5 0

3 years ago