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kati45 [8]
3 years ago
11

3. How many grams of water will be produced from 15.0 grams of Methane? * CH4 +2 02 → CO2 + 2H2O

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
8 0
CH4 + 2O2 = CO2 + 2H2O

According to molar weights :

16 gm CH4 + 64 gm O2 = 44 gm CO2 + 36 gm H2O

Since 16 gm CH4 produce 36 gm H2O

Hence 2.5 gmCH4 produce 36×2.5/16 gm H2O

= 5.265 gm of H2O
2.1K viewsView 2 Upvoters

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A first order reaction has rate constants of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0ºC and 20ºC, respectively. What is the value
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Answer:

D.  18,800 J/mol

Explanation:

We need to use the Arrhenius equation to solve for this problem:

k=Ae^{\frac{-E_a}{RT}, where k is the rate constant, A is the frequency factor, E_a is the activation energy, R is the gas constant, and T is the temperature in Kelvins.

We want to find the value of E_a, so let's plug some of the information we have into the equation. The gas constant we can use here is 8.31 J/mol-K.

At 0°C, which is 0 + 273 = 273 Kelvins, the rate constant k is 4.6*10^{-2}. So:

k=Ae^{\frac{-E_a}{RT}

4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}

At 20°C, which is 20 + 273 = 293 Kelvins, the rate constant k is 8.1*10^{-2}. So:

k=Ae^{\frac{-E_a}{RT}

8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}

We now have two equations and two variables to solve for. We just want to find Ea, so let's write the first equation for A in terms of Ea:

4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}

A=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }

Plug this in for A in the second equation:

8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}

8.1*10^{-2}=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }e^{\frac{-E_a}{8.31*293}

After some troublesome manipulation, the answer should come down to be approximately:

Ea = 18,800 J/mol

The answer is thus D.

5 0
3 years ago
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