Answer:
3,116J/K
Explanation:
This question asks to calculate the entropy change of the surroundings.
To do this, we need the standard enthalpy of formation ΔfH° of the reacting species and products first:
We should observe that standard enthalpy if formation of O2 is zero. We proceed with the rest of the species.
H2CO = -109.5KJ/mol
CO2 = -393.5KJ/mol
H2O = -285.8KJ/mol
Now, we calculate the standard change of enthalpy of the reaction as:
ΔHrxn = ΔHproduct - ΔHreactant = (-285.8 - 393.5) +(109.5) = -569.8 KJ/mol
The relationship between the entropy and the standard formation enthalpy is given as
The relationship is:
ΔSosurroundings = - ( ΔHof/ T)
We convert the standard enthalpy of formation to joules first = -569.8 * 10^3 Joules
Using the formula above at a temperature of 298k, the entropy change would be:
-(-569.8 * 10^3)/298 = 1912J/K
Now, we know that 1.63 moles of H2CO reacted. We also need to know the coefficient of the H2CO in the reaction which is 1.
We thus have:
1.63 mol H2CO(g) * (1912J/K * 1 mol H2CO) = 3116J/K
Just by simply having a water temp.
Answer:
Directly proportional:
A proportion of two variable quantities. when the ratio of two quantities are constant.
inversely proportional:
when one value decreases at the same rate that the other increases.
Explanation:
I hope this does help you, have a great day. :)
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