Assertion : Diamond is thermodynamically less stable than its allotrope graphite
Reason : Molar entropy of graphite is greater than that of diamond under similar conditions.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Statement-1 is True, Statement-2 is False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement-1 is False, Statement-2 is True

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
$C_{(d i a m o n d)} \to C_{(g r a p h i t e)} Δ H_{r} = - v e$
$Δ G_{r x n} = Δ H_{r} - T Δ S$
$Δ S$ is +ve
$∴ Δ G_{r} = - v e \Rightarrow$ spontaneoues reaction
$∴ C_{(g r a p h i t e)}$ is more stable.

Suggest Corrections

Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

N H_{3}

P H_{3}

N H_{3}

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

Join BYJU'S Learning Program