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abruzzese [7]
3 years ago
15

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal

culate the energy change of the reaction at 25 ∘ C and 1.00 bar . Sn ( s ) + 2 Cl 2 ( g ) ⟶ SnCl 4 ( l ) Δ H ∘ rxn = − 511.3 kJ/mol
Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -511.3kJ/mol=-511300.0J/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of \Delta n_gRT = 0

R = gas constant = 8.314 J/mol.K

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

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

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K

\Delta U=-511300.0J/mol+4955.144J/mol

\Delta U=-506344.856J/mol=-506.3kJ/mol

Therefore, the internal energy change is -506.3 kJ/mol

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The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the
deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is 0.3115 \ \text{sec}^{-1}.

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

                                                  k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}

                                               \ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}

where k is the rate constant, E_{a} represents the activation energy of the chemical reaction, R is the gas constant, T is the temperature, and A is the frequency factor.

The frequency factor, A, is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be k_{1} at an initial temperature T_{1} and the rate constant k_{2} at a final temperature T_{2}, thus

                         \ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)

                                         \rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}

Given that E_{a} \ = \ 26.5 \ \ \text{kJ/mol}, R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}, T_{1} \ = \ \left(40 \ + \ 273\right) \ K, T_{2} \ = \ \left(125 \ + \ 273\right) \ K, and k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}, therefore,

           k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}                      

8 0
2 years ago
What is the concentration of chloride ion when 16.8ml of a 0.554 M solution of barium chloride is combined with 16.8ml of a 0.59
ozzi

Answer:

The concentration of chloride ion is 2.887 M.

Explanation:

Here we have a combination BaCl₂  and AlCl₃

1 mole BaCl₂ can produce 2 moles chloride ions as follows;

BaCl₂ → Ba²⁺ + 2Cl⁻

Therefore;

0.554  mole BaCl₂ can produce 2×0.554  moles chloride ions;

1 mole AlCl₃ can produce 3 moles chloride ions as follows;

AlCl₃ → Al³⁺ + 3Cl⁻

Therefore;

0.593 mole AlCl₃ can produce 3×0.593 moles chloride ions

Total concentration of chloride ions = 2×0.554 + 3×0.593 = 2.887 moles of Cl⁻ in 2×16.8 ml solution

The concentration of Cl⁻ is 2.887 M

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3 years ago
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