Question

The standard enthalpy of formation of $C_2H_{4(g)}, CO_{2(g)}$ and $H_2O_{(l)}$ are $52, -394$ and $-286\, kJ\, mol^{-1}$ respectively. Then the amount of heat evolved by burning $7\, g$ of $C_2H_{4(g)}$ is

• A. $1412\, kJ$
• B. $9884\, kJ$
• C. $353\, kJ$
• D. $76\, kJ$

Answer 1

Answer: (C)

Given,
$C+O_2 \rightarrow CO_2 ; Δ Hf = -394\, kJ\, mol^{-1}\,...(i)$
$H_2 +\frac{1}{2}O_2 \rightarrow H_2O\,;ΔH_{f} = -286 \,kJ \,mol^{-1}\, ...(ii)$
$2C+2H_2 \rightarrow C_2H_4\,; Δ Hf = 52\, kJ\, mol^{-1}\, ...(iii)$
Required equation :
$C_2H_4 +3O_2 \rightarrow 2CO_2 +2H_2O\,;ΔH = ?\, ... (iv)$
To get the equation (iv),
multiply equation (i) by $2$, multiply equation (ii) by $2$ and subtract equation (iii)
$(i) \times 2 + (ii) \times 2 - (iii) = iv$
$ΔH = (-3\times 4) \times 2 + (-286) \times 2 - 52 = -1412 \,kJ$
$28\, g$ of $C_2H_4$ gives $1412\, kJ$ heat
$7 \,g \,C_2H_4$ will give $\frac{1412\times7}{28}=353\,kJ$