JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q35

ChemistrySingle correct+4 / -1Question 35 of 90

For the decomposition of the compound, represented as
$\ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)}$
the $K_p=2.9\times10^{-5}$ atm $^3$ .

If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :

1.94\times10^{-2}

atm

5.82\times10^{-2}

atm

7.66\times10^{-2}

atm

38.8\times10^{-2}

atm

Answer and worked solution

Official answer: B

Let $p_{\ce{CO2}}=p$ , so $p_{\ce{NH3}}=2p$ . Then $K_p=(2p)^2p=4p^3=2.9\times10^{-5}$ , giving $p=1.94\times10^{-2}$ atm.

Total: $3p=5.82\times10^{-2}$ atm.

JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 35