Calculate Delta H for the reaction CH4(g)+NH3(g)-->HCN(g)+3H2(g) given,
N2(g)+3H2(g)->2NH3(g) DeltaH=-91.8kj
C(s)+2H2(g)->CH3(g) DeltaH=-74.9kj
h2(g)+2C(s)+N2(g)->2HCN(g) DeltaH=270.3kj
To calculate the enthalpy change (Delta H) for the reaction CH4(g) + NH3(g) -> HCN(g) + 3H2(g), we can use the Hess's law, which states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction.
We can break down the given reaction into three steps:
1. CH4(g) -> C(s) + 2H2(g) (reverse of the third equation)
2. N2(g) + 3H2(g) -> 2NH3(g)
3. C(s) + 2H2(g) + N2(g) -> 2HCN(g) (reverse of the second equation)
Now, we need to assign the appropriate Delta H values to each step and multiply them if necessary:
1. Reverse of the third equation:
Delta H1 = -270.3 kJ (reverse the sign)
2. Second equation:
Delta H2 = -91.8 kJ (given)
3. Reverse of the second equation:
Delta H3 = -(-74.9 kJ) = 74.9 kJ (reverse the sign)
Now, we can add up the Delta H values for each step to find the Delta H for the overall reaction:
Delta H = Delta H1 + Delta H2 + Delta H3
Delta H = -270.3 kJ + (-91.8 kJ) + 74.9 kJ
Delta H ≈ -287.2 kJ
Therefore, the enthalpy change (Delta H) for the reaction CH4(g) + NH3(g) -> HCN(g) + 3H2(g) is approximately -287.2 kJ.
To calculate ΔH for the reaction CH4(g) + NH3(g) → HCN(g) + 3H2(g), you need to use the given ΔH values for the following reactions:
1. N2(g) + 3H2(g) → 2NH3(g) ΔH = -91.8 kJ
2. C(s) + 2H2(g) → CH3(g) ΔH = -74.9 kJ
3. H2(g) + 2C(s) + N2(g) → 2HCN(g) ΔH = 270.3 kJ
The target reaction is not given directly, so we need to manipulate the given reactions in order to obtain the target reaction. The steps to do this are as follows:
1. Reverse reaction 1: 2NH3(g) → N2(g) + 3H2(g)
ΔH = 91.8 kJ (Since the reaction was reversed, the sign of ΔH also changes)
2. Multiply reaction 2 by 3: 3(C(s) + 2H2(g) → CH3(g))
ΔH = -224.7 kJ (Multiply the ΔH value by 3)
Now we can add the manipulated reactions together to obtain the target reaction:
3. 2NH3(g) + 3(C(s) + 2H2(g)) → N2(g) + 3H2(g) + 3CH3(g)
ΔH = 91.8 kJ + (-224.7 kJ)
ΔH = -132.9 kJ
Now, since the target reaction has the same stoichiometry as the manipulated reaction but with different reactants and products, we can directly compare the coefficients of the manipulated reaction with the coefficients of the target reaction:
- Manipulated Reaction: 2NH3(g) + 3(C(s) + 2H2(g)) → N2(g) + 3H2(g) + 3CH3(g)
- Target Reaction: CH4(g) + NH3(g) → HCN(g) + 3H2(g)
By comparing the coefficients, we can see that the manipulated reaction is equivalent to 2 moles of the target reaction. Therefore, we need to divide the ΔH value of the manipulated reaction by 2 to get the ΔH value for the target reaction:
ΔH(target) = -132.9 kJ / 2
ΔH(target) = -66.45 kJ
Therefore, ΔH for the reaction CH4(g) + NH3(g) → HCN(g) + 3H2(g) is -66.45 kJ.