You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Enthalpy of Solution

The reaction ...

Question

The reaction $N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ was carried out at 300K in a bomb
calorimeter. The heat released was $743 k j m o l^{- 1}$ . The value of $△ H_{300 K}$ for this reaction be:-

- 740.5 k j m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 741.75 k j m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 743.0 k j m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 744.25 k j m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $- 740.5 k j m o l^{- 1}$
In a bomb calorimeter, heat released $= - Δ U$
$∴ U = - 743 K J m o l^{- 1}$
$∴ Δ H = Δ U - Δ n g R T$
where, $Δ n g =$ difference between gaseous moles of products and reactants
$∴ Δ n g = (1 + 1) - \frac{3}{2} = \frac{1}{2}$
$∴ Δ H = Δ U + \frac{1}{2} R T$
$= - 743 + \frac{1}{2} \times 8.314 \times 300 \times 10^{- 3}$
$Δ H_{100} = - 741.75 K H m o l^{- 1}$

Suggest Corrections

Similar questions

Q. The reaction

N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)

was carried out at

300 k

in a bomb calorimeter. The heat released was

742 k J m o l^{- 1}

. The value of

△ H_{300 K}

for this reaction would be_________.

Q. The reaction

{N H}_{2} C N (s) + \frac{3}{2} O_{2} (g) + {C O}_{2} (g) + H_{2} O (l)

was carried out at 300 K in a bomb calorimeter. The heat released was

743 k J m o l^{- 1}

. The value of

{Δ H}_{300 K}

for this reaction would be:

{N H}_{2} {C N}_{(s)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + {C O}_{2 (g)} + H_{2} O_{(l)}

This reaction is carried out in a bomb calorie-meter. The heat released was

743 K J {m o l}^{- 1}

. The value of

{Δ H}_{300}

for this reaction would be:

The reaction of cyanamide, NH₂CN_(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol^–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

$N H_{2} C N_{(s)} + \frac{3}{2} O_{2}_{(g)} \to N_{2 (g)} + C O_{2} (g) + H_{2} O_{(l)}$

Q. The reaction of cyanamide

N H_{2} C N (s)

with dioxygen was carried out in a bomb calorimeter and

△ U

was found to be

- 742.7 k J m o l^{- 1}

at 298 K.

N H_{2} C N (g) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)

Calculate enthalpy change for the reaction at 298 K?

Join BYJU'S Learning Program

The reaction NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l) was carried out at 300K in a bombcalorimeter. The heat released was 743 kj mol−1. The value of △H300K for this reaction be:-

The correct option is B −740.5 kj mol−1In a bomb calorimeter, heat released =−ΔU∴U=−743 KJ mol−1∴ΔH=ΔU−ΔngRTwhere, Δng= difference between gaseous moles of products and reactants∴Δng=(1+1)−32=12∴ΔH=ΔU+12RT=−743+12×8.314×300×10−3ΔH100=−741.75 KH mol−1

The reaction $N H_{2} C N (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ was carried out at 300K in a bomb
calorimeter. The heat released was $743 k j m o l^{- 1}$ . The value of $△ H_{300 K}$ for this reaction be:-