Question

question 21 (1 point)
use the following information to answer the next question.
graphite and diamond are two allotropes of carbon. they are formed naturally as well as artificially under different conditions of temperature and pressure. both burn in air, to give carbon dioxide.
(i) c (diamond) + o₂ (g) → co₂ (g) dh = -94.5 kcal
(ii) c (graphite) + o₂ (g) → co₂ (g) dh = -94.0 kcal
what would be the heat of transformation for the following reaction?
c (diamond) → c (graphite)
a) 188.5 kcal
b) -188.5 kcal
c) 0.5 kcal
d) -0.5 kcal
e) -94.5 kcal

Explanation:

Step1: Identify given reactions

We have two reactions:

1. \( \ce{C (diamond) + O2 (g) -> CO2 (g)} \), \( \Delta H_1 = -94.5 \, \text{kCal} \)
2. \( \ce{C (graphite) + O2 (g) -> CO2 (g)} \), \( \Delta H_2 = -94.0 \, \text{kCal} \)

Step2: Manipulate reactions to get target

Target reaction: \( \ce{C (diamond) -> C (graphite)} \)

Subtract reaction 2 from reaction 1 (or reverse reaction 2 and add to reaction 1). Reverse reaction 2: \( \ce{CO2 (g) -> C (graphite) + O2 (g)} \), \( \Delta H_2' = +94.0 \, \text{kCal} \)

Now add reaction 1 and reversed reaction 2:
\( \ce{C (diamond) + O2 (g) + CO2 (g) -> CO2 (g) + C (graphite) + O2 (g)} \)

Simplify to get \( \ce{C (diamond) -> C (graphite)} \)

Step3: Calculate \( \Delta H \)

\( \Delta H = \Delta H_1 + \Delta H_2' = -94.5 + 94.0 = -0.5 \, \text{kCal} \)

Answer:

d) -0.5 KCal