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Rasek [7]
3 years ago
15

To begin the experiment, a bomb calorimeter is filled with 1.11 g CH4 and an excess of oxygen. The heat capacity of the calorime

ter, including the bomb and the water, is 4.319 kJ g C⋅°. The initial temperature of the system was 24.85°C, and the final temperature was 35.65°C.
Using the formula ∆H =−M ⋅ C ⋅∆T , solve for the heat of combustion of. 1.11 g CH4.
Chemistry
1 answer:
Flura [38]3 years ago
6 0
There are several information's already given in the question. Based on those information's the answer can be easily determined.

M = <span>1.11 g CH4
C = </span>4.319 kJ g C⋅°
∆T = 35.65 - 24.85 degree centigrade
     = 10.8 degree centigrade.
Then
∆H =−M ⋅ C ⋅∆T
      = - 1.11 * 4.319 * 10.8
      = - 51.776 kJ/<span>mol

I hope the procedure is clear enough for you to understand.</span>
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Explanation:

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At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9

Now, the new equilibrium condition, taking into account the change x, becomes:

9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}

Thus, we arrange the equation as:

\frac{1}{9} (0.9-2x)^2=(0.2+x)^2\\0.09-0.4x+4x^2=0.04+0.4x+x^2\\3x^2-0.8x+0.05=0\\x_1=0.06

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[NO]_{eq}=0.9-0.06=0.84M

Best regards.

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