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What is the change in enthalpy for the following reaction? 

2H2O2(aq)  2H2O(l) + O2(g) 

Given:
H2O: ∆H= -242 kJ
H2O2: ∆H= -609 kJ

A.

-286 kJ

B. 

-572 kJ

C. 

572 kJ

D. 

286 kJ

User Rob Wagner
by
7.7k points

2 Answers

3 votes

Final answer:

The calculated enthalpy change for the given reaction, using the provided values, is 734 kJ, which does not match the multiple-choice options provided in the question. It suggests that there may be an error in the question or the provided data.

Step-by-step explanation:

The change in enthalpy for the decomposition reaction of hydrogen peroxide (H2O2) into water (H2O) and oxygen (O2) can be calculated using the provided enthalpy values for the reactants and products. To find the enthalpy change for the overall reaction, we subtract the sum of the enthalpy changes of the products (multiplied by their stoichiometric coefficients) from the sum of the enthalpy changes of the reactants (also multiplied by their stoichiometric coefficients).

The given enthalpy of formation for H2O is -242 kJ/mol and for H2O2 is -609 kJ/mol. The equation for the reaction is:

2H2O2(aq) ➔ 2H2O(l) + O2(g)

The enthalpy change (ΔH) for the reaction can be calculated as follows:

ΔH = [ΔH products] - [ΔH reactants] = [(2 mol x -242 kJ/mol) + (1 mol x 0 kJ/mol)] - [2 mol x -609 kJ/mol]

ΔH = [-484 kJ + 0 kJ] - [-1218 kJ]

ΔH = -484 kJ + 1218 kJ = 734 kJ

However, no option closely matches this calculation. It's possible that there has been a misunderstanding or an error in the values or the formulation of the question, as none of the provided choices, A. -286 kJ, B. -572 kJ, C. 572 kJ, D. 286 kJ, correspond to the calculated value of 734 kJ. Therefore, it would be prudent to double-check the values and the phrasing of the question for any mistakes or to seek further clarification.

User Anurag Vohra
by
8.2k points
7 votes

The given chemical reaction is:


2H_(2)O_(2)(aq)---> 2H_(2)O(l)+O_(2)(g)

The standard heats of formation of
H_(2)O and H_(2)O_(2) are:

Δ
H_(f)^(0)(H_(2)O) = -285.8kJ/mol[/tex]

Δ
H_(f)^(0)(H_(2)O_(2)) = -187.6 kJ/mol[/tex]

Calculating the change in heat:

Δ
H^(0)_{reaction)=∑ΔH
_(f)^(0)(products)-∑ΔH
_(f)^(0)(reactants)

= [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]

= -196.4 kJ/mol

Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol



User GauravSingh
by
8.9k points