as 100 milliliters of 0.10 molar KOH is added to 100 milliliters of 0.10 molar HCL at 298 K, the pH of the res?A. decrease to 3B
1 answer:
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<u>Answer:</u>
<u>For 1:</u> The value of for the chemical equation is -24.636 kJ/mol
<u>For 2:</u> The value of for the chemical equation is -20.925 kJ/mol
<u>Explanation:</u>
For the given chemical equation:
- <u>For 1:</u>
To calculate the for given value of equilibrium constant, we use the relation:
.....(1)
where,
= ? kJ/mol
R = Gas constant =
T = temperature = 2000 K
= equilibrium constant in terms of partial pressure = 4.40
Putting values in above equation, we get:
Converting this into kilo joules, we use the conversion factor:
1 kJ = 1000 J
So, -24636.12 J/mol = -24.636 kJ/mol
Hence, the value of for the chemical equation is -24.636 kJ/mol
- <u>For 2:</u>
The expression of for the given chemical equation is:
We are given:
Putting values in above equation, we get:
Now, calculating the value of by using equation 1:
R = Gas constant =
T = temperature = 2000 K
= equilibrium constant in terms of partial pressure = 3.52
Putting values in equation 1, we get:
Converting this into kilo joules, we use the conversion factor:
1 kJ = 1000 J
So, -20925.68 J/mol = -20.925 kJ/mol
Hence, the value of for the chemical equation is -20.925 kJ/mol