Using Hess' Law, how do you calculate the standard heat of formation of Copper(I) Oxide given the following data?
#CuO(s) -> Cu(s)+ 1/2 O_2 # # Delta H = 157.3 kJ# #/# #mol#
#4CuO(s) -> 2Cu_2O(s) + O_2(g)# #Delta H 292.0 kJ# #/# #mol#
You can do it like this:
Hess' Law states that the overall enthalpy change of a process is independent of the route taken.
In thermodynamics we are interested in initial and final states.
You need to construct a Hess Cycle using the information given:
You can see that, in energy terms, the
So we can write:
Enthalpy of formation refers to the formation of 1 mole of a substance from its elements in their standard states under standard conditions. We have found Which is the same as: This refers to the formation of 2 moles of copper(I) oxide. We need the enthalpy change for the formation of 1 mole of copper(I) oxide.
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To calculate the standard heat of formation of Copper(I) Oxide using Hess' Law, you would need the standard heats of formation for the reactants and products involved in the formation reaction. The formation reaction for Copper(I) Oxide is:
Cu(s) + 1/2 O2(g) → Cu2O(s)
Given the standard heats of formation for Cu(s), O2(g), and Cu2O(s), you would use the equation:
ΔH°f(Cu2O) = ΔH°f(Cu(s)) + 1/2 ΔH°f(O2(g)) - ΔH°f(Cu2O(s))
Substitute the given values and solve for ΔH°f(Cu2O).
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To calculate the standard heat of formation of Copper(I) Oxide (Cu2O) using Hess's Law, you need to find a series of reactions involving known standard heats of formation (( \Delta H_f^\circ )) and manipulate them to obtain the desired reaction.
Firstly, you need to find the standard heat of formation of Cu2O. You can use the following reaction:
[ 2Cu(s) + \frac{1}{2}O_2(g) \rightarrow Cu_2O(s) ]
Using known standard heats of formation for Cu(s) and (O_2(g)), you can find ( \Delta H_f^\circ ) for Cu2O.
If you don't have the direct data for Cu2O, you might need to use Hess's Law to derive it from other reactions. This involves manipulating known reactions (with known ( \Delta H_f^\circ ) values) to achieve the desired reaction.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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