1 g of graphite is burnt in a bomb calorimet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The heat absorbed by the calorimeter is given by $q = C_v 	imes \Delta T$.
Given $C_v = 20.7 \ kJ \ K^{-1}$ and $\Delta T = (299 - 298) \ K = 1

Vedclass · Accepted Answer

$-248.4 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The heat absorbed by the calorimeter is given by $q = C_v 	imes \Delta T$.
Given $C_v = 20.7 \ kJ \ K^{-1}$ and $\Delta T = (299 - 298) \ K = 1 \ K$.
Heat released by the reaction $(q_{rxn})$ is equal to $-q_{calorimeter} = -20.7 \ kJ \ K^{-1} 	imes 1 \ K = -20.7 \ kJ$.
This is the internal energy change $(\Delta U)$ for the combustion of $1 \ g$ of graphite.
For $1 \ mol$ of graphite $(12 \ g)$,the internal energy change $(\Delta U)$ is:
$\Delta U = \frac{-20.7 \ kJ}{1 \ g} 	imes 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1}$.
Since the reaction is $C \ (graphite) + O_{2(g)} ightarrow CO_{2(g)}$,the change in moles of gaseous species is $\Delta n_g = 1 - 1 = 0$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,since $\Delta n_g = 0$,we have $\Delta H = \Delta U$.
Therefore,$\Delta H = -248.4 \ kJ \ mol^{-1}$.

Similar Questions

When $200 \ cm^3$ of $0.1 \ M \ H_2SO_4$ is mixed with $150 \ cm^3$ of $0.2 \ M \ KOH$,the heat produced is ..... $kJ$.

$10 \, mol$ of an ideal gas expands isothermally and reversibly from a pressure of $10 \, atm$ to $1 \, atm$ at $300 \, K$. What is the largest mass (in $kg$) which can be lifted through a height of $100 \, m$ by the energy obtained in this process (in $, kg$)?

$\Delta H_f^o$ of water is $-285.5\, kJ\, mol^{-1}$. If enthalpy of neutralisation of monoacidic strong base is $-57.3\, kJ\, mol^{-1}$,$\Delta H_f^o$ of $OH^{-}$ ion will be $.....\, kJ\, mol^{-1}$.

The heats of combustion of $C_2H_4, C_2H_6$ and $H_2$ are $-1409.5 \, kJ, -1558.3 \, kJ$ and $-285.6 \, kJ$ respectively. The heat of hydrogenation of ethene is ...... $kJ$.

The enthalpy of formation of $H_2O$ is $-68 \ k.cal/mol$. Calculate the enthalpy of formation of $OH^-$. Given that the enthalpy of neutralization of $H^+$ and $OH^-$ is $-13.7 \ k.cal/mol$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module