Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (a) to (d) is correct? (i) C(g)+4H(g) → CH4(g); ΔrH = x kJ mol-1 (ii) C(graphite, s)+2H2(g) → CH4(g); ΔrH = y kJ mol-1

Question

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (a) to (d) is correct? (i) C(g)+4H(g) → CH4(g); ΔrH = x kJ mol-1 (ii) C(graphite, s)+2H2(g) → CH4(g); ΔrH = y kJ mol-1 (A) x = y (B) x = 2y (C) x > y (D) x < y

Tardigrade Admin · Accepted Answer

In eqn (i), no bond is being broken while in eqn (ii), 2 H-H bonds are broken. So, in eqn (ii) some of the energy is used up to break the bonds. Thus, x > y.

$x = y$

$x = 2 y$

$x > y$

$x < y$

In eqn $(i)$ , no bond is being broken while in eqn $(ii)$ , $2 H - H$ bonds are broken. So, in eqn $(ii)$ some of the energy is used up to break the bonds. Thus, $x > y$ .

Q. Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (a) to (d) is correct? (i) C(g)​+4H(g)​→CH4(g)​;Δr​H=xkJmol−1 (ii) C(graphite,s)​+2H2(g)​→CH4(g)​;Δr​H=ykJmol−1

x=y

x=2y

x>y

x<y

In eqn (i), no bond is being broken while in eqn (ii), 2H−H bonds are broken. So, in eqn (ii) some of the energy is used up to break the bonds. Thus, x>y.

$x = y$

$x = 2 y$

$x > y$

$x < y$

In eqn $(i)$ , no bond is being broken while in eqn $(ii)$ , $2 H - H$ bonds are broken. So, in eqn $(ii)$ some of the energy is used up to break the bonds. Thus, $x > y$ .