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jeyben [28]
2 years ago
8

2) A common "rule of thumb" -- for many reactions around room temperature is that the

Chemistry
1 answer:
babunello [35]2 years ago
8 0

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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