GATE 2026 BT Q.60: solution and answer

GATE 2026 · BTQ.60 · 2 marks · NAT

A substrate is consumed in a zero order reaction such that its concentration falls from 42 g L $^{- 1}$ to 14 g L $^{- 1}$ in 4 hours. The total time taken for complete utilization of substrate will be ________ hours. (answer in integer)

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Answer and solution

ANSWER

The substrate is fully consumed 6 hours after the start.

Official answer: 6Source: Official answer key, GATE 2026 BT (IIT Guwahati)

Zero-order rate law: $- \frac{d S}{d t} = k_{0}$ , where $S$ is the substrate concentration (in $g L^{- 1}$ ) and $k_{0}$ is the zero-order rate constant (in $g L^{- 1} h^{- 1}$ ). The rate does not depend on $S$ , so integrating from $S_{0}$ at $t = 0$ gives a straight-line fall: $S = S_{0} - k_{0} t$ .
Find $k_{0}$ from the given data ( $S_{0} = 42 g L^{- 1}$ , $S = 14 g L^{- 1}$ at $t = 4 h$ ): $k_{0} = \frac{S _{0} - S}{t} = \frac{42 - 14}{4} = \frac{28}{4} = 7 g L^{- 1} h^{- 1}$ .
Complete utilization means $S = 0$ : $0 = S_{0} - k_{0} t_{complete}$ , so $t_{complete} = \frac{S _{0}}{k _{0}} = \frac{42}{7} = 6 h$ .
Check: the remaining $14 g L^{- 1}$ is consumed at the same constant rate in $14/7 = 2 h$ , and $4 + 2 = 6 h$ in total from the start.

KEY IDEA

In a zero-order reaction the rate is constant, so concentration falls linearly with time: $S = S_{0} - k_{0} t$ . Get $k_{0}$ from the drop you are told about, then complete consumption takes $t = S_{0} / k_{0}$ .

WATCH OUT FOR

Answering 2 h, which is only the time left after the first 4 h, when the question asks for the total time from the start. The other mistake is using first-order kinetics (a logarithmic decay), which never reaches zero concentration and so cannot give a finite time for complete utilization.

From the GATE 2026 BT paper. Every question of this paper · Download the full paper · All 9 Enzyme kinetics questions, with notes

GATE 2026 BT Q.60: solution and answer

Answer and solution

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