Question

In the reaction $3H_2\left(g\right)+N_2\left(g\right)\to 2N{H}_3\left(g\right)$, at constant pressure, the work done by system on surrounding is?

• A. positive
• B. negative
• C. Zero
• D. None of these

Answer 1

The correct option is A positive

$3H_2\left(g\right)+N_2\left(g\right)\to 2N{H}_3\left(g\right)$

Now,

$W=-P_{ext}(V_2-V_1)$

$W=-P_{ext}(n_{2}RT-n_{1}RT)$

$W=-P_{ext}(\Delta n_{g}RT$)

Since $\Delta n_g=-ve$

$\therefore W=+Ve$

Note: $[n_2=2;n_1=3+1=4;\Delta n_g=n_2-n_1=2-4=-2]$