Question

The heat of combustion of ethylene at ${18}^{\circ }C$ and at constant volume is -335.8 K cal when water is obtained in liquid state. What is the heat of combustion at constant pressure and at ${18}^{\circ }C$.

• A. -337 k cal
• B. 337 k cal
• C. 237 k cal
• D. -237k cal

Answer 1

The correct option is A

-337 k cal

The chemical equation for combustion of $C_2{H}_4$ is

$C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right),△U=-335.8Kcal$
Number of moles of reactants = 3 + 1 = 4
Number of moles of products = 2
$△$ n = 2 - 4 = (-2)
Given data $△$ U = -335.8k cal; $△$ n = -2; R = 2 $\times$ 10-3 k cal
T = 18 + 273 = 291K
The change in enthalpy can be given as:
$△H=△U+△nRT$
$△H=-336.964kcal$
$△H\cong -337$ k cal