Suppose you had 1.0 mol samples the following gases at STP. If the volume of each of the samples of gas was reduced to one tenth
of its original size and the temperature remained the same, which gas would have the lowest pressure?
A) He
B)Kr
C)N2
D)NO
E)O2
A) He
B)Kr
C)N2
D)NO
E)O2
2 answers:
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This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
<u>Explanation</u>:
The standard enthalpy change of reaction, Δ H ∘ , is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.
C (s] + O 2(g] → CO 2(g]
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
First to convert grams of carbon into moles,
use carbon's molar mass(12.011 g).
Moles of C = mass in gram / molar mass
= 0.327 g / 12.011 g
Moles of C = 0.027 moles
Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.
= 0.027 moles C 393 kJ
Heat released = 10.611 kJ.
So, when 0.027 moles of carbon react with enough oxygen gas, the reaction will give off 10.611 kJ of heat.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ