Question

The standard enthalples of formation of $CO_2(g)$, $H_2O(l)$ and Isoprene(g) are -400 kJ/mol, -300 kJ/mol and 10 kJ/mol. The standard enthalpy of combustion of Isoprene ($C_5H_8$) at 25°C Is (-t) x $10^4$ J. The value of ($\frac{t}{10}$) Is [Round off to the nearest Integer]

Answer 1

32

Answer 2

The balanced combustion equation for Isoprene ($C_5H_8$) is: $C_5H_8(g) + 7O_2(g) \rightarrow 5CO_2(g) + 4H_2O(l)$

The standard enthalpy of combustion ($\Delta H_c^\circ$) is calculated as: $\Delta H_c^\circ = \sum \nu_p \Delta H_f^\circ(\text{products}) - \sum \nu_r \Delta H_f^\circ(\text{reactants})$

Given: $\Delta H_f^\circ(CO_2(g)) = -400$ kJ/mol $\Delta H_f^\circ(H_2O(l)) = -300$ kJ/mol $\Delta H_f^\circ(C_5H_8(g)) = 10$ kJ/mol $\Delta H_f^\circ(O_2(g)) = 0$ kJ/mol

$\Delta H_c^\circ = [5 \times (-400) + 4 \times (-300)] - [1 \times 10 + 7 \times 0]$ $\Delta H_c^\circ = [-2000 - 1200] - [10]$ $\Delta H_c^\circ = -3200 - 10 = -3210$ kJ/mol

Convert to Joules: $\Delta H_c^\circ = -3210 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -3,210,000$ J

Given that $\Delta H_c^\circ = -t \times 10^4$ J: $-3,210,000 = -t \times 10^4$ $t = \frac{3,210,000}{10,000} = 321$

The value of $\frac{t}{10}$ is $\frac{321}{10} = 32.1$. Rounding to the nearest integer gives 32.