Consider the following data:
C(s) + O2(g) → CO₂(g) AH-394 kJ mol-¹
Pb(s) + 202(g) → PbO(s) AH-217 kJ mol-¹
PbO(s) + CO(g) → Pb(s) + CO2(g) AH-66 kJ mol-¹
Calculate the value of the enthalpy change, in kJ mol, for the following reaction.
C(s) + O2(g) → CO(g)
A-243
B -111
C +111
D +243
The enthalpy change for the reaction C(s) + O2(g) → CO(g) can be calculated by summing the enthalpy changes for the individual reactions and adjusting the sign accordingly.
First, let's write out the chemical equation for the overall reaction:
C(s) + O2(g) → CO(g)
Now, we need to find the corresponding reactions that can add up to the overall reaction. We can see that the first reaction given is already the desired reaction, so we don't need to make any changes to it.
Next, let's look at the second reaction:
Pb(s) + 202(g) → PbO(s) ΔH = -217 kJ mol-¹
We can see that this reaction produces PbO as a product, while we need CO as a product. To do this, we can reverse the second reaction, which will change the sign of the enthalpy change:
PbO(s) → Pb(s) + 202(g) ΔH = +217 kJ mol-¹
Lastly, let's look at the third reaction:
PbO(s) + CO(g) → Pb(s) + CO2(g) ΔH = -66 kJ mol-¹
We can see that this reaction produces CO2 as a product, while we need CO as a product. To do this, we can reverse the third reaction and change the sign of the enthalpy change:
Pb(s) + CO2(g) → PbO(s) + CO(g) ΔH = +66 kJ mol-¹
Now, let's add up the adjusted enthalpy changes for the reactions:
ΔH_total = ΔH_first reaction + ΔH_second reaction + ΔH_third reaction
= -394 kJ mol-¹ + (+217 kJ mol-¹) + (+66 kJ mol-¹)
= -394 + 217 + 66
= -111 kJ mol-¹
So, the value of the enthalpy change, in kJ mol, for the reaction C(s) + O2(g) → CO(g) is -111 kJ mol-¹.
Therefore, the correct answer is B) -111.
To calculate the value of the enthalpy change for the given reaction, we need to sum up the enthalpy changes of the individual reactions involved.
1. We are given:
C(s) + O2(g) → CO2(g) with ΔH = -394 kJ mol-¹
PbO(s) + CO(g) → Pb(s) + CO2(g) with ΔH = -66 kJ mol-¹
2. We can see that the second reaction involves the formation of CO2 as one of the products, which already has an enthalpy change value (-394 kJ mol-¹). Therefore, we need to reverse the given first reaction (C(s) + O2(g) → CO₂(g)) and subtract its enthalpy change from -66 kJ mol-¹ to account for the formation of CO2.
So, the reverse of the first reaction would be: CO2(g) → C(s) + O2(g)
Now, we reverse the sign of the enthalpy change (-394 kJ mol-¹) to account for the reverse reaction.
3. Reversing the sign: CO2(g) → C(s) + O2(g) with ΔH = +394 kJ mol-¹
4. Now, we have:
PbO(s) + CO(g) → Pb(s) + CO2(g) with ΔH = -66 kJ mol-¹
CO2(g) → C(s) + O2(g) with ΔH = +394 kJ mol-¹
5. Add the two reactions together:
PbO(s) + CO(g) + CO2(g) → Pb(s) + C(s) + O2(g)
6. Calculate the overall enthalpy change:
ΔH = (-66 kJ mol-¹) + (+394 kJ mol-¹) = +328 kJ mol-¹
Therefore, the value of the enthalpy change for the given reaction (C(s) + O2(g) → CO(g)) is +328 kJ mol-¹.
Thus, the correct answer is D) +243 kJ mol.
To calculate the value of the enthalpy change for the given reaction, we can use the concept of Hess's Law.
Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
First, let's analyze the given reactions:
1) C(s) + O2(g) → CO2(g) ΔH = -394 kJ mol^(-1)
2) Pb(s) + 2O2(g) → PbO(s) ΔH = -217 kJ mol^(-1)
3) PbO(s) + CO(g) → Pb(s) + CO2(g) ΔH = -66 kJ mol^(-1)
Now, we need to manipulate these reactions to get the desired reaction:
C(s) + O2(g) → CO(g)
By applying some algebraic operations and canceling out the appropriate species, we can manipulate the given reactions to obtain the desired reaction:
1) Multiply reaction 2 by 2:
2Pb(s) + 4O2(g) → 2PbO(s) ΔH = -434 kJ mol^(-1)
2) Flip reaction 2:
2PbO(s) → 2Pb(s) + 4O2(g) ΔH = +434 kJ mol^(-1)
3) Flip reaction 3:
Pb(s) + CO2(g) → PbO(s) + CO(g) ΔH = +66 kJ mol^(-1)
Now, we can add these manipulated reactions together to get the desired reaction:
C(s) + O2(g) + 2PbO(s) → CO(g) + Pb(s) + CO2(g)
Adding up the enthalpy changes of the manipulated reactions:
(-394 kJ mol^(-1)) + (+434 kJ mol^(-1)) + (+66 kJ mol^(-1)) = +106 kJ mol^(-1)
Therefore, the value of the enthalpy change for the given reaction
C(s) + O2(g) → CO(g) is +106 kJ mol^(-1).
Based on the given answer choices, the closest value is +111 kJ mol^(-1), so the correct option would be C: +111.