Chemistry Thermochemistry MDCAT 2012
PMDC Verified Question 77 of 81
Combustion of graphite to form \( CO_2 \) can be done by two ways. Reactions are given as follow:
\( C + O_2 \rightarrow CO_2 \quad \Delta H = -393.7 \text{ kJmol}^{-1} \)
\( C + \frac{1}{2}O_2 \rightarrow CO \quad \Delta H_1 = ? \)
\( CO + \frac{1}{2}O_2 \rightarrow CO_2 \quad \Delta H_2 = -283 \text{ kJmol}^{-1} \)
A
\( -110 \text{ kJ mol}^{-1} \)
B
\( +110 \text{ kJ mol}^{-1} \)
C
\( -676 \text{ kJ mol}^{-1} \)
D
\( 676 \text{ kJ mol}^{-1} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( -110 \text{ kJ mol}^{-1} \)
Concept:

Hess's Law of Constant Heat Summation states that if a reaction can take place by more than one route, the total enthalpy change is identical regardless of the route taken, provided initial and final states are the same.

Formula:

$$ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 $$

Solution:

  • The direct, single-step route is: \( C + O_2 \rightarrow CO_2 \) with \( \Delta H = -393.7 \text{ kJ/mol} \).


  • The two-step indirect route is: Step 1 (forming \( CO \) with \( \Delta H_1 \)) + Step 2 (burning \( CO \) to \( CO_2 \) with \( \Delta H_2 = -283 \text{ kJ/mol} \)).


  • Applying Hess's Law: \( -393.7 = \Delta H_1 + (-283) \).


  • Solve for \( \Delta H_1 \):
    \( \Delta H_1 = -393.7 - (-283) \)
    \( \Delta H_1 = -393.7 + 283 = -110.7 \text{ kJ/mol} \).


  • Rounding to the nearest whole number given in the options yields \( -110 \text{ kJ mol}^{-1} \).


Why other options are incorrect:

Option B has a positive sign, incorrectly implying the partial combustion of carbon is endothermic. Options C and D result from incorrectly adding the two values (\( -393.7 - 283 \)) instead of substituting them properly into the algebraic sum.

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