🛠️ JEE🧪 Chemistry

If \(C\) (diamond) \(\to C\) (graphite)+X \(kJmo{l}^{-1}\) \(C(\text{ diamond })+{O}_{2}(g)\to C{O}_{2}(g)+YkJmo{l}^{-1}…

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If \(C\) (diamond) \(\to C\) (graphite)+X \(kJmo{l}^{-1}\)
\(C(\text{ diamond })+{O}_{2}(g)\to C{O}_{2}(g)+YkJmo{l}^{-1}\)
C (graphite) \(+{O}_{2}(g)\to C{O}_{2}(g)+ZkJmo{l}^{-1}\)

at constant temperature. Then

[JEE Main 2025, 29 Jan (Shift 2)]

a

X = Y - Z

b

X = Y + Z

c

-X=Y+Z

d

X= -Y+Z

✓ Correct answer: a)

X = Y - Z

Explanation

(a) Using Hess's Law, we sum the given reactions:
1. C (diamond) \(\to C\) (graphite) + X
2. C (diamond) \(+{O}_{2}\to C{O}_{2}+Y\)
3. C (graphite) \(+{O}_{2}\to C{O}_{2}+Z\)

Reversing (3):

\(C{O}_{2}\to C(\text{ graphite })+{O}_{2}-Z\)

Adding to (2):
C(diamond) → C(graphite) + (Y–Z)
Comparing with (1): X = Y – Z

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