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3 years ago

Calculate the hfg and sfg of steam at 120oC from the Clapeyron equation, and compare them to the tabulated values.

Chemistry

1 answer:

den301095 [7]3 years ago

5 0

Answer:

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6

kJ/Kg.K

Explanation:

According to Clapeyron Equation :

$h_{fg}=TV_{fg}\left (\frac{\partial P}{\partial T} \right )_{sat}$

$=T(V_{g}-V_{f})_{120^{0}}\left ( \frac{\Delta P}{\125^{0}-115^{0}} \right )$

Taking P2 at temperature 125 and P1 at 115 degree Celsius .

T = 120 + 273.15 = 393.15 K

P2 = 232.23 kPa and P1 = 169.18kPa

(these are experimental values given in the Data)

$= 393.15(0.88133 -0.001060\ m^{3}/Kg)\frac{232.23-169.18 kPa}{10}$

$h_{fg} = 2206.8 kJ/Kg$

$S_{fg}=\frac{h_{fg}}{T}$

$\frac{2206.8}{393.15}$

= 5.6131 kJ/Kg.K

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6131

kJ/Kg.K

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Now we will compare the moles of ammonia with both reactant.

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Calculate the concentration of formate in a 100mm solution of formic acid at ph 4.15. The pka for formic acid is 3.75

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The molarity of formic acid is 100 mM or $100\times 10^{-3}M$ . The dissociation reaction of formic acid is as follows:

$HCOOH\leftrightharpoons HCOO^{-}+H^{+}$

The expression for dissociation constant of the reaction will be:

$K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}$

Rearranging,

$[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}$

Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:

$[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M$

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$[K_{a}=10^{-pK_{a}}=10^{-3.75}=1.78\times 10^{-4}M$

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