Column I Column II a- exothermic p- HP -HR b- endothermic q- - H is negative c- spontaneous r- - H is positive d- heat of reaction s- - G is negative

The correct option is D (a→ q) (b→ r) (c→ s) (d→ p) (a→ q) For exothermic reaction, Δ H= ve (b→ r) For endothermic reac ...

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Byju's Answer

Standard XII

Chemistry

Bond Order and Enthalpy

Column I Col...

Question

Column I Column II

a. exothermic p. $H_{P} - H_{R}$

b. endothermic q. $Δ H$ is negative

c. spontaneous r. $Δ H$ is positive

d. heat of reaction s. $Δ G$ is negative

(a \to s)

(b \to p)

(c \to q)

(d \to r)

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(a \to r)

(b \to s)

(c \to p)

(d \to q)

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(a \to s)

(b \to q)

(c \to p)

(d \to r)

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(a \to q)

(b \to r)

(c \to s)

(d \to p)

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Solution

The correct option is D $(a \to q)$ $(b \to r)$ $(c \to s)$ $(d \to p)$

$(a \to q)$ For exothermic reaction, $Δ H = - v e$
$(b \to r)$ For endothermic reaction, $Δ H = + v e$
$(c \to s)$ For spontaneous process, $Δ G = - v e$ .
$(d \to p)$ $Δ H = H_{P} - H_{R}$

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A car is moving along a straight line. It's displacement (x) - time (t) graph is shown. Match the entries in column I with points on graph as in Column-II.

Column I

(p) x $\to$ negative,v $\to$ positive,a $\to$ positive (q) x $\to$ positive, v $\to$ negative,a $\to$ negative

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Column II

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	Column I		Column II
a.	exothermic	p.	$H_{P} - H_{R}$
b.	endothermic	q.	$Δ H$ is negative
c.	spontaneous	r.	$Δ H$ is positive
d.	heat of reaction	s.	$Δ G$ is negative

Column I Column II a. exothermic p. HP−HR b. endothermic q. ΔH is negative c. spontaneous r. ΔH is positive d. heat of reaction s. ΔG is negative

The correct option is D (a→q) (b→r)(c→s)(d→p)(a→q) For exothermic reaction, ΔH=−ve (b→r) For endothermic reaction, ΔH=+ve(c→s) For spontaneous process, ΔG=−ve.(d→p) ΔH=HP−HR

Column I Column II

a. exothermic p. $H_{P} - H_{R}$

b. endothermic q. $Δ H$ is negative

c. spontaneous r. $Δ H$ is positive

d. heat of reaction s. $Δ G$ is negative

The correct option is D $(a \to q)$ $(b \to r)$ $(c \to s)$ $(d \to p)$

$(a \to q)$ For exothermic reaction, $Δ H = - v e$
$(b \to r)$ For endothermic reaction, $Δ H = + v e$
$(c \to s)$ For spontaneous process, $Δ G = - v e$ .
$(d \to p)$ $Δ H = H_{P} - H_{R}$