Chemical Thermodynamics - SS3 Chemistry Past Questions and Answers - page 4

To find the enthalpy change for the target reaction, we can manipulate the given equations and add them to obtain the desired equation:

1. C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol (Given)

2. 2CO(g) + O2(g) → 2CO2(g) (Multiply Equation 2 by 2)

Now, to make the coefficients of CO2 match in both equations, we can reverse Equation 1:

3. CO2(g) → C(s) + O2(g) ΔH = +393.5 kJ/mol (Reverse Equation 1)

Next, we can add equations 2 and 3 to get the target reaction:

4. 2CO(g) + O2(g) + CO2(g) → 2CO2(g) + C(s) + O2(g)

Now, add the enthalpy changes of the equations to find the enthalpy change for the target reaction:

ΔH_target = ΔH2 + ΔH3

ΔH_target = 0 kJ/mol (Since the products on the right side are the same as the reactants on the left side)

Therefore, the enthalpy change for the reaction: 2CO(g) + O2(g) → 2CO2(g) is 0 kJ/mol.

To find the enthalpy change for the target reaction, we can manipulate the given equations and add them to obtain the desired equation:

1. H2(g) → 2H(g) ΔH = 436.6 kJ/mol (Given)

2. C(s) + 2H(g) → CH4(g) ΔH = -74.8 kJ/mol (Given)

To make the coefficients of H2 match in both equations, we can multiply Equation 2 by 2:

3. 2C(s) + 4H(g) → 2CH4(g) ΔH = -149.6 kJ/mol (Multiply Equation 2 by 2)

Now, add equations 1 and 3 to get the target reaction:

4. H2(g) + 2C(s) + 4H(g) → 2CH4(g) + 2H(g)

Now, add the enthalpy changes of the equations to find the enthalpy change for the target reaction:

ΔH_target = ΔH1 + ΔH3

ΔH_target = 436.6 kJ/mol - 149.6 kJ/mol

ΔH_target = 287.0 kJ/mol

Therefore, the enthalpy change for the reaction: C(s) + 2H2(g) → CH4(g) is 287.0 kJ/mol.