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.
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
1 answer