WhineI - -ve - i - -ve - P - inftyllthline II- infty - ii - 0 - Q - -ve lilhline III- 0 - iii - infty - R - -velllhline iv- -ve - iv - -ve - S - 0 illhline lend array The correct set for melting of ice at -15- C is-A- I iv QB- IV iii PC- I iv RD- IV ii P

The correct option is A (I) (iv) (Q)Normally ice melts above or at 0^∘C. So, if it's melting at 15^∘⇒ non spontaneous. ⇒Δ G =+ve. Due to melting, the entropy ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Enthalpy

WhineI & +ve ...

Question

$\begin{matrix} Δ G & Δ H & Δ S (I) & + v e & (i) & - v e & (P) & \infty (I I) & \infty & (i i) & 0 & (Q) & + v e (I I I) & 0 & (i i i) & \infty & (R) & - v e (i v) & - v e & (i v) & + v e & (S) & 0 \end{matrix}$
The correct set for melting of ice at $- 15^{\circ} C$ is:

(IV) (iii) (P)

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

(I) (iv) (R)

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

(I) (iv) (Q)

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

(IV) (ii) (P)

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

Open in App

Solution

The correct option is C (I) (iv) (Q)
Normally ice melts above or at $0^{\circ} C .$
So, if it's melting at $- 15^{\circ} \Rightarrow$ non spontaneous.
$\Rightarrow Δ G = + v e .$
Due to melting, the entropy is increasing and melting is endothermic
$\Rightarrow Δ S = + v e .$
$\Rightarrow Δ H = + v e .$

Suggest Corrections

Similar questions

\begin{matrix} C o l u m n s 1 r e p r e s e n t s Δ G, & C o l u m n 2 Δ H & a n d C o l u m n 3 Δ S . C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & + v e & (i) & - v e & (P) & \infty (I I) & \infty & (i i) & 0 & (Q) & + v e (I I I) & 0 & (i i i) & \infty & (R) & - v e (I V) & - v e & (i v) & + v e & (S) & 0 \end{matrix}

The correct set for boiling of water at

110^{\circ} C

\begin{matrix} C o l u m n s 1 r e p r e s e n t s Δ G, & C o l u m n 2 Δ H & a n d C o l u m n 3 Δ S . C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & + v e & (i) & - v e & (P) & \infty (I I) & \infty & (i i) & 0 & (Q) & + v e (I I I) & 0 & (i i i) & \infty & (R) & - v e (I V) & - v e & (i v) & + v e & (S) & 0 \end{matrix}

The correct set for Haber’s process is (when equilibrium is just to be reached)

\begin{matrix} List-I & List-II (I) & Work function & (P) & Wavelength of photon used less than threshold wavelength (I I) & Threshold frequency & (Q) & Photo intensity of light (I I I) & Stopping potential & (R) & number of photons (I V) & Photo current & (S) & Metal (T) & Electron (U) & Surface of liquid \end{matrix}

\begin{matrix} Column - I & Column - II i . & [N i (C O)_{4}] & Planar octahedral, diamagnetic i i . & [F e (N O) (H_{2} O)_{5}]^{2 +} & Planar, paramagnetic i i i . & [C o (N O_{2})_{6}]^{3 -} & Octahedral, diamagnetic i v . & [C u (N H_{3})_{4}]^{2 +} & Planar octahedral, paramagnetic Octahedral, paramagnetic \end{matrix}

Q. Match the Column-X with Column-Y:
Given reaction:

A + B_{2} \to A B_{2}

\begin{matrix} Column X & Column Y (i) 300 atoms of A + 200 molecules of B_{2} & (p) B_{2} is limiting reagent (ii) 100 atoms of A + 100 molecules of B_{2} & (q) A is limiting reagent (iii) 5 mol of A + 2.5 mol of B_{2} & (r) None of the reactant is in excess (iv) 2.5 mol of A + 5 mol of B_{2} & (s) 200 molecules of A B_{2} will be formed \end{matrix}

Join BYJU'S Learning Program

ΔGΔHΔS(I)+ve(i)−ve(P)∞(II)∞(ii)0(Q)+ve(III)0(iii)∞(R)−ve(iv)−ve(iv)+ve(S)0 The correct set for melting of ice at −15∘C is:

The correct option is C (I) (iv) (Q)Normally ice melts above or at 0∘C. So, if it's melting at −15∘⇒ non spontaneous. ⇒ΔG=+ve. Due to melting, the entropy is increasing and melting is endothermic ⇒ΔS=+ve. ⇒ΔH=+ve.

$\begin{matrix} Δ G & Δ H & Δ S (I) & + v e & (i) & - v e & (P) & \infty (I I) & \infty & (i i) & 0 & (Q) & + v e (I I I) & 0 & (i i i) & \infty & (R) & - v e (i v) & - v e & (i v) & + v e & (S) & 0 \end{matrix}$
The correct set for melting of ice at $- 15^{\circ} C$ is: