Question

1. (a) 0.504 g of H2 reacts with O2 at constant pressure evolving 17.08 kcal. Calculate
enthalpy change for the reaction
H2(g) +1/2O2 (g) > H20 (l)​

Answer 1

We know that, ΔCp = ΣnCp products - ΣnCp reactants = [(1x 75.312) - ( 1x 38.83 - 1/2 x 29.16 )] = 51.06 J/K/mol = 0.051 kJ/K

Further , we know from Kirchoff's law,

ΔH(at final temperature) –ΔH(at initial temperature) =[ (Cp(final) -Cp (initial) (T(final ) -T( initial)] = ΔCp (Tf -Ti )

Or, ΔH ( 373K) - ΔH (298K) = ΔCp (Tf -Ti )

Or, ΔH ( 373K)- (-285.76 kJ/K) = 0.051kJ/K x ( 373K-298K) = 3.825 kJ

or, ΔH ( 373K) = 3.825 - 285.76 = -281.93 kJ

Hence, ΔH ( at 373K) = -281.93 kJ

Answer 2

Answer:

The enthalpy change for the reaction is 68.32 kcal/mol.

Explanation:

The balanced equation for the reaction is:

2H2(g) + O2(g) → 2H2O(l)

The enthalpy change for this reaction can be calculated using the following formula:

ΔH = q/n

where:

   ΔH is the enthalpy change (in kcal/mol)

   q is the heat released or absorbed during the reaction (in kcal)

   n is the number of moles of the limiting reactant

First, we need to calculate the number of moles of H2 involved in the reaction:

n(H2) = 0.504 g / (2.016 g/mol) = 0.25 mol

The enthalpy change can then be calculated as follows:

ΔH = 17.08 kcal / 0.25 mol = 68.32 kcal/mol

Therefore, the enthalpy change for the reaction is 68.32 kcal/mol.

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