Consider a reaction -N2-3H2 - 2NH3Kp for the above reaction at 127-C is 4-34 - 10-3 -r H - -460-6 cal mol-1 - Estimate the temperature at which Kp will be 43-4 - 10-3

Van't Hoff Equation : log(K_2K_1)= Δ H^∘2.303R(1T_2 1T_1) Given T_1=127^∘C=(127+273)K=400K Δ H^∘= 460.6 cal mol^ 1 K_1=4.34 × 10^ 3 K_2=43.4 × 10^ 3 Also, R=2 ...

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

Byju's Answer

Standard XII

Chemistry

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

Consider a re...

Question

Consider a reaction :
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$K_{p}$ for the above reaction at $127^{\circ} C$ is $4.34 \times 10^{- 3}$ .
$Δ_{r} H =$ $- 460.6 c a l m o l^{- 1}$ . Estimate the temperature at which $K_{p}$ will be $43.4 \times 10^{- 3}$

150 K

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

200 K

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

30 K

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

80 K

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

Open in App

Solution

The correct option is D $80 K$
Van't Hoff Equation : $log (\frac{K_{2}}{K_{1}}) = \frac{- Δ H^{\circ}}{2.303 R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
Given $T_{1} = 127^{\circ} C = (127 + 273) K = 400 K Δ H^{\circ} = - 460.6 c a l m o l^{- 1} K_{1} = 4.34 \times 10^{- 3} K_{2} = 43.4 \times 10^{- 3}$
Also, $R = 2 c a l K^{- 1} m o l^{- 1}$
Putting the values:
$log (\frac{43.4 \times 10^{- 3}}{4.34 \times 10^{- 3}}) = \frac{+ 460.6}{2.303 \times 2} (\frac{1}{T_{2}} - \frac{1}{400})$
$l o g (10) = 100 \times (\frac{1}{T_{2}} - \frac{1}{400}) \frac{1}{100} = (\frac{1}{T_{2}} - \frac{1}{400}) \frac{1}{T_{2}} = \frac{1}{100} + \frac{1}{400} = \frac{1}{80} T_{2} = 80 K$

Suggest Corrections

Similar questions

Q. Consider a reaction :

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K_{p}

for the above reaction at

127^{\circ} C

4.34 \times 10^{- 3}

Δ_{r} H =

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

. Estimate the temperature at which

K_{p}

will be

43.4 \times 10^{- 3}

K_{p}

for the reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

1.6 \times 10^{- 4} a t m^{- 2}

400^{\circ} C

. What will be

K_{p}

500^{\circ} C

? Heat of reaction in this temperature range is

- 25.14 k c a l

K_{p}

for the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

400^{\circ} C

1.64 \times 10^{- 4}

Calculate

K_{C}

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressures 2 atm,1 atm, and 3 atm respectively at 725K.

If the value of

K_{p}

for the reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

4.28 \times 10^{- 5} {a t m}^{- 2}

at 725K,in which direction the net reaction will go :

K_{P}

for the reaction

N_{2} + {2 H}_{2} ⇌ {2 N H}_{3}

1.6 \times 10^{- 4} {a t m}^{- 2}

400^{0} C

. What will be

K_{P}

500^{0} C

? The heat of reaction at this temperature is

- 25.14

kcal?

Join BYJU'S Learning Program

Consider a reaction : N2+3H2⇌2NH3 Kp for the above reaction at 127∘C is 4.34×10−3 . ΔrH= −460.6 cal mol−1 . Estimate the temperature at which Kp will be 43.4×10−3

Consider a reaction :
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$K_{p}$ for the above reaction at $127^{\circ} C$ is $4.34 \times 10^{- 3}$ .
$Δ_{r} H =$ $- 460.6 c a l m o l^{- 1}$ . Estimate the temperature at which $K_{p}$ will be $43.4 \times 10^{- 3}$