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1 mol of nitrogen is mixed with 3 mol of...

`1` mol of nitrogen is mixed with `3` mol of hydrogen in a `4 L` container. If `0.25%` of nitrogen is converted to ammonia by the following reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
then calculate the equilibrium constant `K_(c)` in concentration units. What will be the value of `K_(c)` for the following equilibrium?
`1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)`

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The correct Answer is:
A, C
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One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) , then calculate the equilibrium constant, K_(c) in concentration, units. What will be the value of K_(c) for the following equilibrium? 1/2 N_(2)(g)+3/2H_(2)(g)hArrNH_(3)(g)

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Knowledge Check

  • 1 mol of N_(2) is mixed with 3 mol of H_(2) in a litre container. If 50% of N_(2) is converted into ammonia by the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , then the total number of moles of gas at the equilibrium are

    A
    `1.5`
    B
    `4.5`
    C
    `3.0`
    D
    `6.0`

  • 1 mole of nitrogen is mixed with 3 moles of hydrogen in a sqrt3 litre container where 66.67% of nitrogen is converted into ammonia by the following reaction : N_2(g)+3H_2(g) hArr 2NH_3(g) ,Then the value of K_C for the reaction NH_3(g) hArr 1/2N_2(g) + 3/2H_2 will be :

    A
    2 M
    B
    `1/2`M
    C
    4 M
    D
    `1/4` M

  • One mole of nitrogen is mixed with 3 mole of hydrogen in a closed 3 litre vessel. 20% of nitrogen is converted into NH_(3) . Then K_(C) for the (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)hArr NH_(3) is

    A
    0.36
    B
    0.46
    C
    0.5
    D
    0.2