How many liters of CO(g) at STP are produced when 68.0 g of CaCO3(s) is heated according to the following equation? CaCO3(s) CaO
2 answers:
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Explanation:
The given data is as follows.
m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)
Initial temp. of block 1, = (100 + 273) K = 373 K
Initial temp. of block 2, = (0 + 273) K = 273 K
So, heat released by block 1 = heat gained by block 2
Convert temperature into kelvin as (50 + 273) K = 323 K.
Also, we know that the relation between enthalpy and temperature change is as follows.
=
= 1243550 J
or, = 1243.5 kJ
Now, calculate entropy change for block 1 as follows.
=
=
= -554.12 J/K
Now, entropy change for block 2 is as follows.
=
=
= 647.49 J/K
Hence, total entropy will be sum of entropy change of both the blocks.
= -554.12 J/K + 647.49 J/K
= 93.37 J/K
Thus, we can conclude that for the given reaction is 1243.5 kJ and
is 93.37 J/K.