Home
Class 12
CHEMISTRY
For the reaction N(2)(g) + 3H(2)(g) rarr...

For the reaction `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) , DeltaH = -93.6 KJ mol^(-1)` the formation of `NH_(3)` is expected to increase at :

A

High pressure and low temperature

B

Low pressure and low temperature

C

High pressure and high temperature

D

Low pressure and high temperature

Text Solution

Verified by Experts

The correct Answer is:
A
On increasing pressure, equilibrium shifts in the direction where number of moles of gaseous reactant or product is less. Thus high pressure and low temperature is favorable condition.
Promotional Banner

Similar Questions

Explore conceptually related problems

For the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?

For the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g),DeltaH=-93.6 kJ mol^(-1) . The concentration of H_(2) at equilibrium will increase if

For the reaction N_(2)(G)+3H_(2)(g)hArr2NH_(3)(g),Delta=-93.6KJmol^(-1) The number of moles o fH_(2) at equilibrium will increase If :

In reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH=-93.6kJ /mole, the yield of ammonia does not increase when

For the reaction : N_(2(g))+3H_(2(g))hArr2NH_(3(g)) , DeltaH=-ve , the correct statement is :

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(2)(g) Which of the following is correct?

Given N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g),Delta_(r)H^(Ө)= -92.4 kJ mol^(-1) What is the standard enthalpy of formation of NH_(3) gas?

Given: N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g), Delta_(r)H^(@) = -924 kJ "mol"^(-1) . What is the standard enthalpy of formation of NH_(3) gas?

For the reaction, N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g) DeltaH =- 95.0 kJ and DeltaS = - 19000 J K^(-1) Calculate the temperature in centigrade at which it will attain equilibrium.