hoose any two temperatures and corresponding vapor pressures in the table given, and use those values to calculate for dichloromethane (in ).

Question

hoose any two temperatures and corresponding vapor pressures in the table given, and use those values to calculate for dichloromethane (in ).

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263 80.1
273 133.6
283 213.3
293 329.6
303 495.4
313 724.4

Answer 1

Your question makes no sense.

Answer 2

Choose any two temperatures and corresponding vapor pressures in the table given, and use those values to calculate change in vaporization for dichloromethane (in kj/mol).

T(K) (Pvap (mmHg)
263 80.1
273 133.6
283 213.3
293 329.6
303 495.4
313 724.4

Answer 3

To calculate the enthalpy change for dichloromethane (CH2Cl2) at a given temperature, you can use the Clausius-Clapeyron equation:

ln(P2/P1) = (∆H/R) * (1/T1 - 1/T2)

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
∆H is the enthalpy change.
R is the ideal gas constant (8.314 J/(mol·K)).
T1 and T2 are the corresponding temperatures in Kelvin.

Let's choose two temperatures from the given table and the corresponding vapor pressures:

Temperature 1 (T1): 283 K
Vapor pressure at T1 (P1): 213.3 mmHg

Temperature 2 (T2): 293 K
Vapor pressure at T2 (P2): 329.6 mmHg

Now we can substitute these values into the Clausius-Clapeyron equation:

ln(329.6/213.3) = (∆H/8.314) * (1/283 - 1/293)

First, we can simplify the equation:

ln(1.544) = (∆H/8.314) * (0.003535 - 0.003415)

Next, let's solve for (∆H/8.314) by dividing both sides by (0.003535 - 0.003415):

ln(1.544) / (0.003535 - 0.003415) = ∆H/8.314

Now, calculate the left side of the equation:

ln(1.544) / 0.00012 ≈ 42.656 ≈ ∆H/8.314

Multiply both sides by 8.314 to get ∆H:

∆H ≈ 354.617 J/mol

So, the approximate enthalpy change (∆H) for dichloromethane at these temperatures is 354.617 J/mol.