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likoan [24]
2 years ago
12

Help pls do the last two all work and last two questions

Chemistry
1 answer:
andrew11 [14]2 years ago
8 0

d1 = 15 miles

t1 = 1 hour

d2 = 25 miles

t2 = 2 hours

d = 15 + 25 = 40 miles

t = 1 + 2 = 3 hours

avg. \: speed \:  =  \:  \frac{total \: distance}{total \: time}

avg. \: speed \:  =  \frac{40}{3}

avg. \: speed \:  = 13.33mi \: hr ^{ - 1}

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8 0
3 years ago
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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}}

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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

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\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

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