2 g of carbon is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure. During the reaction, temperature rises from 298 K to 300 K. If the heat capacity of the bomb calorimeter is 20.7 kJ / K, what is the enthalpy change for the above reaction at 298 K and 1 atm ? C (graphite) + O 2( g ) arrow CO 2( g )

Question

2 g of carbon is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure. During the reaction, temperature rises from 298 K to 300 K. If the heat capacity of the bomb calorimeter is 20.7 kJ / K, what is the enthalpy change for the above reaction at 298 K and 1 atm ? C (graphite) + O 2( g ) arrow CO 2( g ) (A) -20.7 kJ / mol (B) -2.48 × 102 kJ / mol (C) -41.4 kJ / mol (D) -4.96 × 102 kJ / mol

Tardigrade Admin · Accepted Answer

As we know that, q =- C V × Δ T . =-20.7 ×(300-298)=-41.4 kJ For combustion of 1 mol of graphite =(12.0 g / mol ×(-41.4)/1) =-4.96 × 102 kJ / mol , text since Δ n g =0 Δ H =Δ E =-4.96 × 102 kJ / mol

$- 20.7 k J / m o l$

$- 2.48 \times 1 0^{2} k J / m o l$

$- 41.4 k J / m o l$

$- 4.96 \times 1 0^{2} k J / m o l$

As we know that, $q = - C_{V} \times Δ T$
. $= - 20.7 \times (300 - 298) = - 41.4 k J$
For combustion of $1 m o l$ of graphite
$= \frac{12.0 g / m o l \times ( - 41.4 )}{1}$
$= - 4.96 \times 1 0^{2} k J / m o l, since Δ n_{g} = 0$
$Δ H = Δ E = - 4.96 \times 1 0^{2} k J / m o l$

−20.7kJ/mol

−2.48×102kJ/mol

−41.4kJ/mol

−4.96×102kJ/mol

As we know that, q=−CV​×ΔT . =−20.7×(300−298)=−41.4kJ For combustion of 1mol of graphite =112.0g/mol×(−41.4)​ =−4.96×102kJ/mol, since Δng​=0 ΔH=ΔE=−4.96×102kJ/mol

$- 20.7 k J / m o l$

$- 2.48 \times 1 0^{2} k J / m o l$

$- 41.4 k J / m o l$

$- 4.96 \times 1 0^{2} k J / m o l$

As we know that, $q = - C_{V} \times Δ T$
. $= - 20.7 \times (300 - 298) = - 41.4 k J$
For combustion of $1 m o l$ of graphite
$= \frac{12.0 g / m o l \times ( - 41.4 )}{1}$
$= - 4.96 \times 1 0^{2} k J / m o l, since Δ n_{g} = 0$
$Δ H = Δ E = - 4.96 \times 1 0^{2} k J / m o l$