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ipn [44]
3 years ago
9

Methane burns in oxygen to produce carbon dioxide and water. Which of the tollowing represents a preliminary, unbalanced equatio

n with the correct chemical formulas?
Chemistry
1 answer:
WITCHER [35]3 years ago
4 0

So first we need to know what the formulas are:

Methane: CH_{4}

Oxygen: O_{2}

Carbon Dioxide: CO_{2}

Water: H_{2}O

So then we follow the equation as specified in the question. Note that it does not need to be balanced, as the question states it wants the unbalanced equation:

CH_{4}+O_{2} → CO_{2}+H_{2}O

This is the preliminary, unbalanced equation with the correct chemical formulas.

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

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To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

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K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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