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ehidna [41]
3 years ago
10

Who performed the oil-drop experiment?

Chemistry
2 answers:
evablogger [386]3 years ago
7 0

In the year 1909, Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.

faltersainse [42]3 years ago
7 0

Answer:

Robert Millikan

Explanation:

(APEX)

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Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
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<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

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