You visited us 1 times! Enjoying our articles? Unlock Full Access!

Valency and Oxidation State

Calculate the...

Question

Calculate the temperature at which $Δ_{r} H$ for the reaction will be zero.
Given : $C_{p . m} (C H_{4}, g) = 38 J / K m o l$
$C_{p . m} (C O, g) = 31 J / K m o l$
$C_{p . m} (C H_{3} C H O . g) = 52 J K / m o l$ ]\

1300^{0}

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

1027^{0}

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

700^{0}

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

754^{0}

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

Open in App

Solution

The correct option is C $1300^{0}$ C
$△ H_{2} = - 17 K J / m o l$
$T_{2} = 300 K$
$△ H_{1} = 0$
$T_{1} = ?$
$C_{p} = 17 / m o l$
Using the reaction, $(△ H_{2} - △ H_{1}) = C_{p} (T_{2} - T_{1})$
$- 17.0 K J / m o l - 0 = (17 J / K m o l) (300 K - T_{1})$
$\frac{- 17 \times 1000 J / m o l}{17 J / K m o l} = 300 K - T_{1}$
$T_{1} = 300 K + 1000 K$
$T_{1} = 1300 K$

Suggest Corrections

Similar questions

C_{p . m} = a T^{3},

C_{p . m}

J / K m o l

10 K

10 K

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K_{p}

127^{\circ} C

4.34 \times 10^{- 3}

Δ_{r} H =

- 460.6 c a l m o l^{- 1}

K_{p}

43.4 \times 10^{- 3}

Δ_{r} H

Δ_{r} S

M g O (s) + C (s) \to M g (s) + C O (g)

491.18

m o l^{- 1}

197.67

m o l^{- 1}

K_{p}

A (g) ⇌ B (g) + C (g)

27 C a n d 4 a t 47 .

B (g) + C (g) ⇌ A (g)

(G i v e n : R = 2 c a l / m o l - K)

Δ H

85^{o} C

{F e}_{2} O_{3} (s) + 3 H_{2} ⟶ 2 F e (s) + 3 H_{2} O (l)

Δ H_{298}^{o} = - 33.29 k J / m o l

Substance	${F e}_{2} O_{3} (s)$	$F e (s)$	$H_{2} O (l)$	$H_{2} (g)$
$C_{P}^{o} (J / K m o l)$	$103.8$	$25.1$	$75.3$	$28.8$

Join BYJU'S Learning Program