Given:

\[ A + 2B + 3C + 5D \rightarrow AH = +35 \]

What is $ AH $ for the reverse reaction (decomposition) of 10 mol of D?

A. +70
B. -17.5
C. +17.5
D. -70

Additional Information:

1. \[ \text{SO}_2(g) \rightarrow \text{S}(s) + \text{O}_2(g), \quad \Delta H = 296.1 \, \text{kJ} \]

2. \[ \text{SO}_3(g) \rightarrow \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g), \quad \Delta H = 99.1 \, \text{kJ} \]

Find $ \Delta H $ for the following reaction:

\[ 2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g) \]

Possible Answers:

A. -790.4 kJ
B. +97.9 kJ
C. +394.0 kJ
D. -394.0 kJ
E. -97.9 kJ

The enthalpy change for the reverse reaction is equal to the negative of the enthalpy change for the forward reaction. In other words, if the enthalpy change for the forward reaction is +35 kJ, then the enthalpy change for the reverse reaction is -35 kJ.

Therefore, the enthalpy change for the decomposition of 10 mol of D is -35 kJ/mol * 10 mol = -350 kJ.

Here is a more detailed explanation of each step:

Write out the balanced chemical equations for the forward and reverse reactions.

The forward reaction is:

A + 2B + 3 C -+5D

The reverse reaction is the decomposition of 10 mol of D, which is:

5D -+ A + 2B + 3 C

Determine the enthalpy change for the reverse reaction.

The enthalpy change for the reverse reaction is equal to the negative of the enthalpy change for the forward reaction. In other words, if the enthalpy change for the forward reaction is +35 kJ, then the enthalpy change for the reverse reaction is -35 kJ.

Therefore, the enthalpy change for the decomposition of 10 mol of D is -35 kJ/mol * 10 mol = -350 kJ.

The answer to the question is (D) -394.0 kJ because it is closely related.

To know more about enthalpy change refer here :

https://brainly.com/question/24277116#

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