P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:
1 answer:
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Answer : The enthalpy of reaction is, -96.9 kJ/mole
Explanation :
First we have to calculate the mass of solution.
Volume of solution = Volume of HCl + Volume of
Volume of solution = 56.6 mL + 36.5 mL
Volume of solution = 93.1 mL
Density of solution = 1 g/mL
The mass of solution is, 93.1 grams.
Now we have to calculate the heat released in the system.
Formula used :
or,
where,
Q = heat released = ?
m = mass = 93.1 g
= specific heat capacity of water =
= initial temperature =
= final temperature =
Now put all the given value in the above formula, we get:
(1 kJ = 1000 J)
Now we have to calculate the moles of and HCl.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 2 mole of
So, moles of
react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 1 mole of react to give 2 mole of
So, moles of
react with
moles of
Now we have to calculate the change in enthalpy of the reaction.
where,
= enthalpy of reaction = ?
q = heat of reaction = 1.88 kJ
n = moles of reaction = 0.0194 mole
Now put all the given values in above expression, we get:
The negative sign indicates that the heat is released.
Therefore, the enthalpy of reaction is, -96.9 kJ/mole