In calculating the change in enthalpy in a certain reaction, it is important to remember that it is the sum of the enthapy of formation for the substances involved in the reaction.
<span>CaO(s)+2HCl(aq)→CaCl2(aq)+H2O(l) ΔH= –186kJ
</span>Ca(OH)2(aq) → CaO(s)+H2O(l)<span> ΔH= 65.1 kJ
</span>Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) ΔH= -120.9 kJ
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
Answer:
There is 17.1 kJ energy required
Explanation:
Step 1: Data given
Mass of ethanol = 322.0 grams
Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K
Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K
Specific heat capacity = 2.44 J/g*K
Step 2: Calculate energy
Q = m*c*ΔT
⇒ m = the mass of ethanol= 322 grams
⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K
⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K
Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ
There is 17.1 kJ energy required
<span>2H2 + O2 → 2H2O</span>
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<span>okay???</span>
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