For complete combustion of methanol $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ the amount of heat produced as measured by bomb calorimeter is $726 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$. The enthalpy of combustion for the reaction is $-\mathrm{x} \mathrm{kJ} \mathrm{mol}^{-1}$, where x is \_\_\_\_. (Nearest integer) (Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Chemical Thermodynamics

For complete combustion of methanol

CH_{3} OH (l) + \frac{3}{2} O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l)

the amount of heat produced as measured by bomb calorimeter is

726 kJ mol^{- 1}

2 7^{\circ} C

. The enthalpy of combustion for the reaction is

- x kJ mol^{- 1}

, where x is ____. (Nearest integer)
(Given :

R = 8.3 JK^{- 1} mol^{- 1}

)

Answer: 727

The bomb calorimeter measures internal energy change:

Δ U = - 726

kJ/mol. To find enthalpy change, use the relation

Δ H = Δ U + Δ n_{g} R T

, where

Δ n_{g}

is the change in moles of gas.

Δ n_{g} = n_{products} - n_{reactants} = 1 - \frac{3}{2} = - \frac{1}{2}

Δ H = - 726 + (- \frac{1}{2}) \times 8.3 \times 1 0^{- 3} \times 300

Δ H = - 726 - 1.245 = - 727.245

kJ/mol

Rounding to the nearest integer:

x = 727

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer